OP Malhotra Class 11 Maths Solutions Chapter 23 Parabola Exercise 23

Question 1. The focus at (10, 0) the directrix x = -10.
Answer: The given focus is at \( (10, 0) \) and the directrix is \( x = -10 \). Since the focus is on the x-axis and the directrix is perpendicular to the x-axis, the axis of the parabola lies along the x-axis. We know the standard equation of such a parabola is \( y^2 = 4ax \). Here, the distance from the origin to the focus is \( a = 10 \). Therefore, we substitute this value into the equation.
The required equation of the parabola is \( y^2 = 4(10)x \), which simplifies to \( y^2 = 40x \). This parabola opens towards the positive x-axis.
In simple words: The parabola has its special point (focus) at (10, 0) and a guiding line (directrix) at x = -10. So, its equation is \( y^2 = 40x \).

🎯 Exam Tip: Remember that if the focus is \( (a, 0) \) and the directrix is \( x = -a \), the parabola opens to the right and its equation is \( y^2 = 4ax \). Always identify 'a' correctly based on the given focus and directrix.

Question 2. The focus at (0, 5), the directrix y = − 5.
Answer: The given focus is at \( (0, 5) \) and the directrix is \( y = -5 \). Since the focus is on the y-axis and the directrix is parallel to the x-axis, the axis of the parabola lies along the y-axis. Comparing the focus \( (0, 5) \) with \( (0, a) \), we find that \( a = 5 \). The standard equation for such a parabola (opening upwards) is \( x^2 = 4ay \). We substitute the value of \( a = 5 \) into the equation.
The required equation of the parabola is \( x^2 = 4(5)y \), which simplifies to \( x^2 = 20y \). This parabola opens upwards. This means its vertex is at the origin.
In simple words: The parabola's special point (focus) is at (0, 5) and its guiding line (directrix) is at y = -5. So, its equation is \( x^2 = 20y \).

🎯 Exam Tip: When the focus is \( (0, a) \) and the directrix is \( y = -a \), the parabola opens upwards. Its equation is \( x^2 = 4ay \). If the focus were \( (0, -a) \) and directrix \( y = a \), it would open downwards, \( x^2 = -4ay \).

Question 3. The focus at (- 3,0), the directrix x + 5 = 0.
Answer: Let \( P(x, y) \) be any point on the parabola. The focus \( F \) is at \( (-3, 0) \) and the directrix is the line \( x + 5 = 0 \). By the definition of a parabola, the distance from any point \( P \) on the parabola to the focus \( F \) is equal to its perpendicular distance to the directrix. This means \( |PF| = |PM| \).
So, we write the distances: \( \sqrt{(x - (-3))^2 + (y - 0)^2} = \frac{|x + 5|}{\sqrt{1^2 + 0^2}} \).
This simplifies to \( \sqrt{(x + 3)^2 + y^2} = |x + 5| \).
Now, we square both sides to remove the square root: \( (x + 3)^2 + y^2 = (x + 5)^2 \).
Expanding both sides, we get: \( x^2 + 6x + 9 + y^2 = x^2 + 10x + 25 \).
Subtracting \( x^2 \) from both sides and rearranging terms: \( y^2 = 10x - 6x + 25 - 9 \).
This results in \( y^2 = 4x + 16 \). We can factor out 4 to get \( y^2 = 4(x + 4) \). This is the required equation of the parabola. This parabola opens to the right.
In simple words: For any point on a parabola, its distance to a fixed point (focus) is the same as its distance to a fixed line (directrix). Using this rule for focus \( (-3, 0) \) and directrix \( x+5=0 \), we find the parabola's equation to be \( y^2 = 4(x+4) \).

🎯 Exam Tip: When the focus is not at the origin or the directrix is not a simple axis, always use the definition \( |PF| = |PM| \). Ensure you correctly apply the distance formula for a point to a line \( \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} \).

Question 4. The focus at (2, – 3), the directrix x + 5 = 0.
Answer: Let \( P(x, y) \) be any point on the parabola. The focus \( F \) is at \( (2, -3) \) and the directrix is the line \( x + 5 = 0 \). According to the definition of a parabola, the distance from \( P \) to \( F \) must be equal to the perpendicular distance from \( P \) to the directrix. So, \( |PF| = |PM| \).
We write the distance equations: \( \sqrt{(x - 2)^2 + (y - (-3))^2} = \frac{|x + 5|}{\sqrt{1^2 + 0^2}} \).
This simplifies to \( \sqrt{(x - 2)^2 + (y + 3)^2} = |x + 5| \).
Next, we square both sides: \( (x - 2)^2 + (y + 3)^2 = (x + 5)^2 \).
Expanding the terms: \( (x^2 - 4x + 4) + (y^2 + 6y + 9) = (x^2 + 10x + 25) \).
Combine like terms: \( x^2 - 4x + 4 + y^2 + 6y + 9 = x^2 + 10x + 25 \).
Subtract \( x^2 \) from both sides and move terms to one side: \( y^2 + 6y + 13 - 4x = 10x + 25 \).
Rearranging to find the equation of the parabola: \( y^2 + 6y - 14x - 12 = 0 \). This is the required equation of the parabola. This parabola has its axis parallel to the x-axis.
In simple words: We find the equation for a parabola using its focus \( (2, -3) \) and directrix \( x+5=0 \). By setting the distance from a point \( (x,y) \) to the focus equal to its distance to the directrix, we get the equation \( y^2 - 14x + 6y - 12 = 0 \).

🎯 Exam Tip: When the directrix equation is \( Ax + By + C = 0 \), always remember that the denominator in the distance formula is \( \sqrt{A^2 + B^2} \). For a directrix like \( x+5=0 \), \( A=1, B=0 \), so the denominator is \( \sqrt{1^2+0^2} = 1 \).

Question 5. The focus at (1, 1), the directrix x – y = 3.
Answer: Let \( P(x, y) \) be any point on the parabola. The focus \( F \) is at \( (1, 1) \) and the directrix is the line \( x - y - 3 = 0 \). According to the definition of a parabola, \( |PF| = |PM| \).
So, we set up the distance equation: \( \sqrt{(x - 1)^2 + (y - 1)^2} = \frac{|x - y - 3|}{\sqrt{1^2 + (-1)^2}} \).
This simplifies to \( \sqrt{(x - 1)^2 + (y - 1)^2} = \frac{|x - y - 3|}{\sqrt{2}} \).
Now, we square both sides: \( (x - 1)^2 + (y - 1)^2 = \frac{(x - y - 3)^2}{2} \).
Multiply both sides by 2: \( 2[(x - 1)^2 + (y - 1)^2] = (x - y - 3)^2 \).
Expand both sides: \( 2(x^2 - 2x + 1 + y^2 - 2y + 1) = x^2 + y^2 + 9 - 2xy - 6x + 6y \).
Simplify further: \( 2x^2 + 2y^2 - 4x - 4y + 4 = x^2 + y^2 + 9 - 2xy - 6x + 6y \).
Move all terms to one side to get the equation of the parabola: \( 2x^2 - x^2 + 2y^2 - y^2 + 2xy - 4x + 6x - 4y - 6y + 4 - 9 = 0 \).
This gives us: \( x^2 + y^2 + 2xy + 2x - 10y - 5 = 0 \). This is the required equation of the parabola. The presence of the \( 2xy \) term shows its axis is not parallel to either coordinate axis.
In simple words: We find the parabola's equation where its focus is \( (1, 1) \) and its directrix is \( x-y-3=0 \). Using the rule that a point on the parabola is equally distant from the focus and directrix, we square and simplify to get \( x^2 + y^2 + 2xy + 2x - 10y - 5 = 0 \).

🎯 Exam Tip: Be very careful when expanding \( (x - y - 3)^2 \). Use the identity \( (a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca \) to avoid errors, remembering that \( b=-y \) and \( c=-3 \).

Question 6. The vertex at the origin, the axis along the x-axis, and passes through (- 3, 6).
Answer: We are given that the vertex of the parabola is at the origin \( (0,0) \) and its axis is along the x-axis. The standard equation for such a parabola is \( y^2 = 4ax \) (if it opens right) or \( y^2 = -4ax \) (if it opens left).
The parabola passes through the point \( (-3, 6) \). We substitute these coordinates into the equation \( y^2 = 4ax \).
\( (6)^2 = 4a(-3) \).
\( 36 = -12a \).
To find \( a \), we divide 36 by -12: \( a = \frac{36}{-12} = -3 \).
Since \( a \) is negative, the parabola opens towards the negative x-axis, so the form \( y^2 = 4ax \) is correct and simply yields a negative \( a \).
Now, substitute \( a = -3 \) back into the standard equation: \( y^2 = 4(-3)x \).
The required equation of the parabola is \( y^2 = -12x \). This parabola opens to the left.
In simple words: The parabola starts at \( (0,0) \), opens left or right along the x-axis, and goes through point \( (-3,6) \). By putting the point into the equation \( y^2 = 4ax \), we find \( a=-3 \), so the final equation is \( y^2 = -12x \).

🎯 Exam Tip: When the vertex is at the origin and the axis is along the x-axis, the equation is \( y^2 = 4ax \). The sign of \( a \) determines if it opens right (positive \( a \)) or left (negative \( a \)). Similarly, for the y-axis, \( x^2 = 4ay \).

Question 7. The focus at (-2, -1) and the latus rectum joins the points (-2, 2) and (- 2, – 4)
Answer: Given the focus is \( F(-2, -1) \) and the endpoints of the latus rectum are \( (-2, 2) \) and \( (-2, -4) \).
Since the x-coordinates of both endpoints of the latus rectum are the same (which is -2), this means the latus rectum is a vertical line. This implies that the axis of the parabola is parallel to the x-axis.
The length of the latus rectum is the distance between its endpoints: \( \sqrt{(-2 - (-2))^2 + (2 - (-4))^2} = \sqrt{0^2 + (2 + 4)^2} = \sqrt{0 + 6^2} = \sqrt{36} = 6 \).
The length of the latus rectum is also equal to \( 4a \). So, \( 4a = 6 \implies a = \frac{6}{4} = \frac{3}{2} \).
The focus is \( (-2, -1) \). For a parabola with its axis parallel to the x-axis, the vertex is \( (h, k) \) and the focus is \( (h+a, k) \) or \( (h-a, k) \).
Given the focus is \( (-2, -1) \), and the latus rectum is vertical, the parabola opens either left or right.
If the parabola opens right, focus is \( (h+a, k) \). So, \( h+a = -2 \) and \( k = -1 \).
\( h + \frac{3}{2} = -2 \implies h = -2 - \frac{3}{2} = -\frac{4}{2} - \frac{3}{2} = -\frac{7}{2} \).
The vertex is \( (-\frac{7}{2}, -1) \). The equation is \( (y - k)^2 = 4a(x - h) \).
\( (y - (-1))^2 = 4(\frac{3}{2})(x - (-\frac{7}{2})) \)
\( (y + 1)^2 = 6(x + \frac{7}{2}) \)
\( (y + 1)^2 = 6(\frac{2x + 7}{2}) \)
\( (y + 1)^2 = 3(2x + 7) \)
\( y^2 + 2y + 1 = 6x + 21 \)
\( y^2 + 2y - 6x - 20 = 0 \).

If the parabola opens left, focus is \( (h-a, k) \). So, \( h-a = -2 \) and \( k = -1 \).
\( h - \frac{3}{2} = -2 \implies h = -2 + \frac{3}{2} = -\frac{4}{2} + \frac{3}{2} = -\frac{1}{2} \).
The vertex is \( (-\frac{1}{2}, -1) \). The equation is \( (y - k)^2 = -4a(x - h) \).
\( (y - (-1))^2 = -4(\frac{3}{2})(x - (-\frac{1}{2})) \)
\( (y + 1)^2 = -6(x + \frac{1}{2}) \)
\( (y + 1)^2 = -6(\frac{2x + 1}{2}) \)
\( (y + 1)^2 = -3(2x + 1) \)
\( y^2 + 2y + 1 = -6x - 3 \)
\( y^2 + 2y + 6x + 4 = 0 \).
The source shows only these two cases from some intermediate steps, I will follow the source's logic to obtain the same final equations.
In simple words: We found the length of the latus rectum using the given points, which helped us find \( a \). Because the latus rectum is vertical, the parabola opens sideways. We considered two possibilities: it opens to the right or to the left, and for each case, we calculated the vertex and the parabola's equation.

🎯 Exam Tip: When given the endpoints of the latus rectum and the focus, first determine if the parabola's axis is parallel to the x-axis or y-axis. The length of the latus rectum is \( 4a \). Based on the focus's position relative to the latus rectum's center, you can determine the vertex and the direction of opening.

Question 8. The vertex at (- 2,3) and the focus at (1,3).
Answer: Given the vertex is \( V(-2, 3) \) and the focus is \( F(1, 3) \).
Notice that the y-coordinates of the vertex and focus are the same (both are 3). This means the axis of the parabola is a horizontal line \( y = 3 \), which is parallel to the x-axis.
For such a parabola, the standard equation is \( (y - k)^2 = 4a(x - h) \), where \( (h, k) \) is the vertex. Here \( (h, k) = (-2, 3) \).
The distance \( a \) is the distance between the vertex and the focus. So, \( a = \sqrt{(1 - (-2))^2 + (3 - 3)^2} = \sqrt{(1 + 2)^2 + 0^2} = \sqrt{3^2} = 3 \).
Since the focus \( (1, 3) \) is to the right of the vertex \( (-2, 3) \), the parabola opens to the right.
Substitute \( h = -2 \), \( k = 3 \), and \( a = 3 \) into the standard equation:
\( (y - 3)^2 = 4(3)(x - (-2)) \).
\( (y - 3)^2 = 12(x + 2) \).
Expand the equation:
\( y^2 - 6y + 9 = 12x + 24 \).
Rearrange the terms to get the general form of the parabola's equation:
\( y^2 - 6y - 12x + 9 - 24 = 0 \).
\( y^2 - 6y - 12x - 15 = 0 \). This is the required equation of the parabola.
In simple words: The parabola's starting point (vertex) is \( (-2,3) \) and its special point (focus) is \( (1,3) \). Since both have the same 'y' value, the parabola opens sideways. We calculate the distance between them as \( a=3 \). Using the formula for a parabola opening right, we get the equation \( y^2 - 6y - 12x - 15 = 0 \).

🎯 Exam Tip: When the vertex and focus have the same y-coordinate, the axis is horizontal and the parabola opens left or right. The value of 'a' is the distance between the vertex and focus, and its sign determines the opening direction (positive 'a' for right, negative 'a' for left).

Question 9. The vertex at (0, 0) and the focus at (0, 1).
Answer: Given the vertex is \( V(0, 0) \) and the focus is \( F(0, 1) \).
Since the x-coordinates of both the vertex and focus are the same (both are 0), the axis of the parabola lies along the y-axis. The focus \( (0, 1) \) is above the vertex \( (0, 0) \), which means the parabola opens upwards.
The standard equation for a parabola with vertex at the origin and axis along the y-axis is \( x^2 = 4ay \).
The distance \( a \) is the distance between the vertex and the focus: \( a = \sqrt{(0 - 0)^2 + (1 - 0)^2} = \sqrt{0^2 + 1^2} = 1 \).
Now, substitute \( a = 1 \) into the standard equation:
\( x^2 = 4(1)y \).
The required equation of the parabola is \( x^2 = 4y \). This parabola opens upwards.
In simple words: The parabola starts at \( (0,0) \) and its special point is \( (0,1) \). This means it opens upwards along the y-axis. The distance from the start to the special point is \( a=1 \). So, the equation is \( x^2 = 4y \).

🎯 Exam Tip: If the vertex is at the origin and the focus is \( (0, a) \), the parabola's axis is the y-axis, and it opens upwards if \( a > 0 \). The equation will always be \( x^2 = 4ay \).

Question 10. The vertex at (0, a) and the focus at (0,0).
Answer: Given the vertex is \( V(0, a) \) and the focus is \( F(0, 0) \).
Since the x-coordinates of both the vertex and focus are the same (both are 0), the axis of the parabola lies along the y-axis. The focus \( (0, 0) \) is below the vertex \( (0, a) \) (assuming \( a > 0 \)). This means the parabola opens downwards.
The standard equation for a parabola with vertex at \( (h, k) \) and axis along the y-axis, opening downwards, is \( (x - h)^2 = -4p(y - k) \).
Here, the vertex is \( (h, k) = (0, a) \). The distance \( p \) (often represented as \( a \) in general forms, distinct from the coordinate 'a' here) between the vertex and the focus is \( p = \sqrt{(0 - 0)^2 + (0 - a)^2} = \sqrt{(-a)^2} = |a| \). Assuming \( a \) is a positive constant for the vertex coordinate.
Let's use \( p \) for the focal length to avoid confusion. So, \( p = a \).
Substitute \( h = 0 \), \( k = a \), and \( p = a \) into the equation:
\( (x - 0)^2 = -4a(y - a) \).
The required equation of the parabola is \( x^2 = -4a(y - a) \). This parabola opens downwards.
In simple words: The parabola's vertex is at \( (0, a) \) and its special point (focus) is at \( (0,0) \). Since the focus is below the vertex, the parabola opens downwards along the y-axis. The equation turns out to be \( x^2 = -4a(y - a) \).

🎯 Exam Tip: If the vertex is \( (h, k) \) and the axis is parallel to the y-axis, the general equation is \( (x - h)^2 = \pm 4p(y - k) \). The sign depends on whether it opens up (+) or down (-). Remember to correctly identify \( p \) as the focal length.

Question 11. The axis parallel to the x-axis, and the parabola passes through (3, 3), (6, 5) and (6, – 3).
Answer: Since the axis of the parabola is parallel to the x-axis, its general equation can be taken as \( x = ay^2 + by + c \), where \( a, b, c \) are constants. We are given that the parabola passes through three points: \( (3, 3) \), \( (6, 5) \), and \( (6, -3) \). We will substitute these points into the equation to form a system of linear equations.

For point \( (3, 3) \):
\( 3 = a(3)^2 + b(3) + c \)
\( 3 = 9a + 3b + c \) ...(Equation 1)

For point \( (6, 5) \):
\( 6 = a(5)^2 + b(5) + c \)
\( 6 = 25a + 5b + c \) ...(Equation 2)

For point \( (6, -3) \):
\( 6 = a(-3)^2 + b(-3) + c \)
\( 6 = 9a - 3b + c \) ...(Equation 3)

Now, we solve these equations simultaneously.
Subtract Equation 3 from Equation 1:
\( (3 - 6) = (9a + 3b + c) - (9a - 3b + c) \)
\( -3 = 9a - 9a + 3b - (-3b) + c - c \)
\( -3 = 6b \)
\( b = \frac{-3}{6} = -\frac{1}{2} \).

Now, substitute \( b = -\frac{1}{2} \) into Equation 1 and Equation 2.
From Equation 1:
\( 3 = 9a + 3(-\frac{1}{2}) + c \)
\( 3 = 9a - \frac{3}{2} + c \)
\( 3 + \frac{3}{2} = 9a + c \)
\( \frac{6}{2} + \frac{3}{2} = 9a + c \)
\( \frac{9}{2} = 9a + c \) ...(Equation 4)

From Equation 2:
\( 6 = 25a + 5(-\frac{1}{2}) + c \)
\( 6 = 25a - \frac{5}{2} + c \)
\( 6 + \frac{5}{2} = 25a + c \)
\( \frac{12}{2} + \frac{5}{2} = 25a + c \)
\( \frac{17}{2} = 25a + c \) ...(Equation 5)

Now, subtract Equation 4 from Equation 5:
\( (\frac{17}{2} - \frac{9}{2}) = (25a + c) - (9a + c) \)
\( \frac{8}{2} = 25a - 9a \)
\( 4 = 16a \)
\( a = \frac{4}{16} = \frac{1}{4} \).

Finally, substitute \( a = \frac{1}{4} \) into Equation 4 to find \( c \):
\( \frac{9}{2} = 9(\frac{1}{4}) + c \)
\( \frac{9}{2} = \frac{9}{4} + c \)
\( c = \frac{9}{2} - \frac{9}{4} \)
\( c = \frac{18}{4} - \frac{9}{4} \)
\( c = \frac{9}{4} \).

Now substitute the values of \( a = \frac{1}{4} \), \( b = -\frac{1}{2} \), and \( c = \frac{9}{4} \) back into the general equation \( x = ay^2 + by + c \).
\( x = \frac{1}{4}y^2 - \frac{1}{2}y + \frac{9}{4} \).
To clear the fractions, multiply the entire equation by 4:
\( 4x = y^2 - 2y + 9 \).
Rearrange the terms to get the required equation of the parabola:
\( y^2 - 2y - 4x + 9 = 0 \). This is the required equation of the parabola.
In simple words: Since the parabola's axis is parallel to the x-axis, its equation looks like \( x = ay^2 + by + c \). We used the three given points to create three equations. Solving them helped us find the values for \( a, b, c \). Putting these values back into the equation gave us \( y^2 - 2y - 4x + 9 = 0 \).

🎯 Exam Tip: When a parabola passes through three general points, assume the appropriate general equation (e.g., \( x = ay^2 + by + c \) for axis parallel to x-axis, or \( y = ax^2 + bx + c \) for axis parallel to y-axis). Substitute the points to form a system of equations and solve for the coefficients.

Question 12. The axis parallel to the y-axis and the parabola passes through the points (4, 5), (-2, 11) and (-4, 21).
Answer: Since the axis of the parabola is parallel to the y-axis, its general equation can be taken as \( y = Ax^2 + Bx + C \), where \( A, B, C \) are arbitrary constants. The parabola passes through three given points: \( (4, 5) \), \( (-2, 11) \), and \( (-4, 21) \). We substitute these points into the equation.

For point \( (4, 5) \):
\( 5 = A(4)^2 + B(4) + C \)
\( 5 = 16A + 4B + C \) ...(Equation 1)

For point \( (-2, 11) \):
\( 11 = A(-2)^2 + B(-2) + C \)
\( 11 = 4A - 2B + C \) ...(Equation 2)

For point \( (-4, 21) \):
\( 21 = A(-4)^2 + B(-4) + C \)
\( 21 = 16A - 4B + C \) ...(Equation 3)

Now, we solve these equations simultaneously.
Subtract Equation 1 from Equation 3:
\( (21 - 5) = (16A - 4B + C) - (16A + 4B + C) \)
\( 16 = 16A - 16A - 4B - 4B + C - C \)
\( 16 = -8B \)
\( B = \frac{16}{-8} = -2 \).

Now, substitute \( B = -2 \) into Equation 1 and Equation 2.
From Equation 1:
\( 5 = 16A + 4(-2) + C \)
\( 5 = 16A - 8 + C \)
\( 5 + 8 = 16A + C \)
\( 13 = 16A + C \) ...(Equation 4)

From Equation 2:
\( 11 = 4A - 2(-2) + C \)
\( 11 = 4A + 4 + C \)
\( 11 - 4 = 4A + C \)
\( 7 = 4A + C \) ...(Equation 5)

Now, subtract Equation 5 from Equation 4:
\( (13 - 7) = (16A + C) - (4A + C) \)
\( 6 = 16A - 4A \)
\( 6 = 12A \)
\( A = \frac{6}{12} = \frac{1}{2} \).

Finally, substitute \( A = \frac{1}{2} \) into Equation 5 to find \( C \):
\( 7 = 4(\frac{1}{2}) + C \)
\( 7 = 2 + C \)
\( C = 7 - 2 = 5 \).

Now substitute the values of \( A = \frac{1}{2} \), \( B = -2 \), and \( C = 5 \) back into the general equation \( y = Ax^2 + Bx + C \).
\( y = \frac{1}{2}x^2 - 2x + 5 \).
To clear the fraction, multiply the entire equation by 2:
\( 2y = x^2 - 4x + 10 \).
Rearrange the terms to get the required equation of the parabola:
\( x^2 - 4x - 2y + 10 = 0 \). This is the required equation of the parabola.
In simple words: Since the parabola's axis is parallel to the y-axis, its equation is \( y = Ax^2 + Bx + C \). We used the three points it passes through to create three equations. By solving these equations for \( A, B, C \), we found the final equation of the parabola to be \( x^2 - 4x - 2y + 10 = 0 \).

🎯 Exam Tip: Always set up a system of three linear equations for the unknown constants \( A, B, C \) when a curve passes through three distinct points. Careful algebraic manipulation is key to solving these systems accurately.

Question 13. The parabola y² = 4px passes through the point (3, – 2). Obtain the length of the latus rectum and the coordinates of the focus.
Answer: The given equation of the parabola is \( y^2 = 4px \). This is the standard form of a parabola with its vertex at the origin and axis along the x-axis. The parabola passes through the point \( (3, -2) \). We substitute these coordinates into the equation to find \( p \).
\( (-2)^2 = 4p(3) \)
\( 4 = 12p \)
\( p = \frac{4}{12} = \frac{1}{3} \).

Now we can find the length of the latus rectum and the coordinates of the focus.
The length of the latus rectum for a parabola of the form \( y^2 = 4px \) is \( 4p \).
Length of latus rectum \( = 4p = 4(\frac{1}{3}) = \frac{4}{3} \).

The coordinates of the focus for a parabola of the form \( y^2 = 4px \) are \( (p, 0) \).
Coordinates of the focus \( = (\frac{1}{3}, 0) \).
In simple words: We are given the parabola's equation \( y^2 = 4px \) and a point \( (3, -2) \) it passes through. By putting the point into the equation, we find \( p = \frac{1}{3} \). Then, the length of the latus rectum is \( 4p = \frac{4}{3} \), and the focus is at \( (p, 0) = (\frac{1}{3}, 0) \).

🎯 Exam Tip: For a parabola in the form \( y^2 = 4px \), always remember that the focus is \( (p, 0) \) and the length of the latus rectum is \( 4p \). If it were \( x^2 = 4py \), the focus would be \( (0, p) \) and the latus rectum \( 4p \).

Question 14. Prove that the equation y² + 2ax + 2by + c = 0 represents a parabola whose axis is parallel to the axis of x. Find its vertex.
Answer: The given equation is \( y^2 + 2ax + 2by + c = 0 \). To prove that this represents a parabola with its axis parallel to the x-axis, we need to rewrite it in the standard form \( (y - k)^2 = 4p(x - h) \).
First, group the terms involving \( y \) and move the other terms to the right side:
\( y^2 + 2by = -2ax - c \).
Next, complete the square for the y-terms. To do this, add \( (b)^2 \) to both sides:
\( y^2 + 2by + b^2 = -2ax - c + b^2 \).
The left side is now a perfect square: \( (y + b)^2 = -2ax + b^2 - c \).
Factor out \( -2a \) from the terms on the right side. This step is crucial to get the \( (x-h) \) form:
\( (y + b)^2 = -2a \left[ x - \frac{b^2 - c}{2a} \right] \).
This equation is in the form \( (y - k)^2 = 4p(x - h) \), where:
\( k = -b \)
\( 4p = -2a \implies p = -\frac{a}{2} \)
\( h = \frac{b^2 - c}{2a} \)
Since the equation can be transformed into the standard form \( (y - k)^2 = 4p(x - h) \), it represents a parabola. The form itself indicates that the axis of symmetry is parallel to the x-axis (it's \( y = k \)).
The vertex of this parabola is \( (h, k) \).
So, the vertex is \( \left( \frac{b^2 - c}{2a}, -b \right) \). This successfully proves that the given equation represents a parabola with its axis parallel to the x-axis and provides its vertex. The focal length depends on the value of 'a'.
In simple words: We take the given equation \( y^2 + 2ax + 2by + c = 0 \) and rearrange it by completing the square for the \( y \) terms. This changes it into the form \( (y+b)^2 = -2a(x - \frac{b^2-c}{2a}) \). This new form clearly shows it's a parabola that opens sideways (its axis is parallel to the x-axis). The starting point, or vertex, is \( (\frac{b^2-c}{2a}, -b) \).

🎯 Exam Tip: To transform a general quadratic equation into the standard form of a parabola, always start by completing the square for the squared variable's terms. Then, factor out the coefficient of the linear variable on the other side to reveal the vertex and focal length.

Question 15. Of the parabola, 4(y − 1)² = − 7 (x – 3) find (i) the length of the latus rectum. (ii) the coordinates of the focus and the vertex.
Answer: The given equation of the parabola is \( 4(y - 1)^2 = -7(x - 3) \). To find the latus rectum, focus, and vertex, we need to convert this into the standard form \( (y - k)^2 = -4a(x - h) \).
Divide both sides by 4:
\( (y - 1)^2 = -\frac{7}{4}(x - 3) \).

Comparing this to the standard form \( (y - k)^2 = -4a(x - h) \), we can identify the following:
\( k = 1 \)
\( h = 3 \)
\( -4a = -\frac{7}{4} \).

(i) **Length of the latus rectum:**
From \( -4a = -\frac{7}{4} \), we get \( 4a = \frac{7}{4} \). The length of the latus rectum is \( |4a| \).
Length of latus rectum \( = \frac{7}{4} \).

(ii) **Coordinates of the focus and the vertex:**
The vertex of the parabola is \( (h, k) \).
Vertex \( = (3, 1) \).

Since the parabola is of the form \( (y - k)^2 = -4a(x - h) \), it opens towards the negative x-axis (to the left). The focus for such a parabola is at \( (h - a, k) \).
First, find \( a \): \( 4a = \frac{7}{4} \implies a = \frac{7}{16} \).
Now, calculate the focus:
Focus \( = (h - a, k) = (3 - \frac{7}{16}, 1) \).
\( 3 - \frac{7}{16} = \frac{3 \times 16}{16} - \frac{7}{16} = \frac{48}{16} - \frac{7}{16} = \frac{41}{16} \).
So, the coordinates of the focus are \( (\frac{41}{16}, 1) \).
In simple words: We changed the given parabola equation into a standard form. From this, we found that the latus rectum's length is \( \frac{7}{4} \). The vertex (starting point) of the parabola is \( (3,1) \). Since the parabola opens to the left, we subtracted \( a \) from the x-coordinate of the vertex to find the focus, which is \( (\frac{41}{16}, 1) \).

🎯 Exam Tip: Always convert the given equation into a standard form, such as \( (y-k)^2 = 4a(x-h) \) or \( (x-h)^2 = 4a(y-k) \), to easily identify the vertex \( (h,k) \) and the value of \( 4a \) (latus rectum length). The sign of \( 4a \) indicates the direction of opening and helps determine the focus coordinates.

Question 16. Find the vertex, focus, and directrix of the following parabolas :
(i) y² – 2y + 8x – 23 = 0
(ii) x² + 8x + 12y + 4 = 0
Answer:
(i) Given equation: \( y^2 - 2y + 8x - 23 = 0 \).
To find the vertex, focus, and directrix, we complete the square for the \( y \) terms and move the \( x \) terms to the other side.
\( y^2 - 2y = -8x + 23 \).
Add \( (-1)^2 = 1 \) to both sides to complete the square for \( y \):
\( y^2 - 2y + 1 = -8x + 23 + 1 \).
\( (y - 1)^2 = -8x + 24 \).
Factor out -8 from the right side:
\( (y - 1)^2 = -8(x - 3) \).
This is in the standard form \( (y - k)^2 = -4a(x - h) \).
Comparing, we get: \( k = 1 \), \( h = 3 \), and \( -4a = -8 \implies 4a = 8 \implies a = 2 \).

**Vertex:** \( (h, k) = (3, 1) \).
Since it's \( (y - k)^2 = -4a(x - h) \) and \( a > 0 \), the parabola opens to the left.

**Focus:** For a parabola opening left, the focus is at \( (h - a, k) \).
Focus \( = (3 - 2, 1) = (1, 1) \).

**Directrix:** For a parabola opening left, the directrix is the line \( x = h + a \).
Directrix \( = x = 3 + 2 \implies x = 5 \).

(ii) Given equation: \( x^2 + 8x + 12y + 4 = 0 \).
To find the vertex, focus, and directrix, we complete the square for the \( x \) terms and move the \( y \) terms to the other side.
\( x^2 + 8x = -12y - 4 \).
Add \( (\frac{8}{2})^2 = 4^2 = 16 \) to both sides to complete the square for \( x \):
\( x^2 + 8x + 16 = -12y - 4 + 16 \).
\( (x + 4)^2 = -12y + 12 \).
Factor out -12 from the right side:
\( (x + 4)^2 = -12(y - 1) \).
This is in the standard form \( (x - h)^2 = -4a(y - k) \).
Comparing, we get: \( h = -4 \), \( k = 1 \), and \( -4a = -12 \implies 4a = 12 \implies a = 3 \).

**Vertex:** \( (h, k) = (-4, 1) \).
Since it's \( (x - h)^2 = -4a(y - k) \) and \( a > 0 \), the parabola opens downwards.

**Focus:** For a parabola opening downwards, the focus is at \( (h, k - a) \).
Focus \( = (-4, 1 - 3) = (-4, -2) \).

**Directrix:** For a parabola opening downwards, the directrix is the line \( y = k + a \).
Directrix \( = y = 1 + 3 \implies y = 4 \).
In simple words: For both parabolas, we first changed their equations into a standard form by completing the square. Once in standard form, we could easily read off the vertex (the parabola's turning point), calculate the focus (the special point), and determine the equation of the directrix (the guiding line).

🎯 Exam Tip: Always remember the different standard forms: \( (y-k)^2 = \pm 4a(x-h) \) for horizontal axis and \( (x-h)^2 = \pm 4a(y-k) \) for vertical axis. The sign determines the direction of opening, which is crucial for finding the correct focus and directrix formulas.

Question 17. Find the equation to the parabola whose axis is parallel to the y-axis and which passes through the points (0, 4), (1, 9) and (- 2, 6) and determine its latus rectum.
Answer: Since the axis of the parabola is parallel to the y-axis, its general equation can be written as \( y = ax^2 + bx + c \). The parabola passes through the points \( (0, 4) \), \( (1, 9) \), and \( (-2, 6) \). We will substitute these points to form a system of equations.

For point \( (0, 4) \):
\( 4 = a(0)^2 + b(0) + c \)
\( 4 = c \).

For point \( (1, 9) \):
\( 9 = a(1)^2 + b(1) + c \)
\( 9 = a + b + c \). Since \( c=4 \):
\( 9 = a + b + 4 \)
\( a + b = 5 \) ...(Equation 1)

For point \( (-2, 6) \):
\( 6 = a(-2)^2 + b(-2) + c \)
\( 6 = 4a - 2b + c \). Since \( c=4 \):
\( 6 = 4a - 2b + 4 \)
\( 2 = 4a - 2b \)
Divide by 2: \( 1 = 2a - b \) ...(Equation 2)

Now we solve Equation 1 and Equation 2 for \( a \) and \( b \).
Add Equation 1 and Equation 2:
\( (a + b) + (2a - b) = 5 + 1 \)
\( 3a = 6 \)
\( a = \frac{6}{3} = 2 \).

Substitute \( a = 2 \) into Equation 1:
\( 2 + b = 5 \)
\( b = 5 - 2 = 3 \).

So, we have \( a = 2 \), \( b = 3 \), and \( c = 4 \). Substitute these values back into the general equation \( y = ax^2 + bx + c \).
The equation of the parabola is \( y = 2x^2 + 3x + 4 \).

Now we need to determine its latus rectum. To do this, we rewrite the equation in the standard form \( (x - h)^2 = 4p(y - k) \).
\( y = 2x^2 + 3x + 4 \)
Divide by 2: \( \frac{y}{2} = x^2 + \frac{3}{2}x + 2 \).
Complete the square for the x-terms. Add \( (\frac{3/2}{2})^2 = (\frac{3}{4})^2 = \frac{9}{16} \) to both sides:
\( \frac{y}{2} = x^2 + \frac{3}{2}x + \frac{9}{16} + 2 - \frac{9}{16} \).
\( \frac{y}{2} = (x + \frac{3}{4})^2 + \frac{32 - 9}{16} \).
\( \frac{y}{2} = (x + \frac{3}{4})^2 + \frac{23}{16} \).
Move the constant term to the left:
\( \frac{y}{2} - \frac{23}{16} = (x + \frac{3}{4})^2 \).
\( (x + \frac{3}{4})^2 = \frac{8y - 23}{16} \).
\( (x + \frac{3}{4})^2 = \frac{8}{16}(y - \frac{23}{8}) \).
\( (x + \frac{3}{4})^2 = \frac{1}{2}(y - \frac{23}{8}) \).
Comparing this to \( (x - h)^2 = 4p(y - k) \), we see that \( 4p = \frac{1}{2} \).
The length of the latus rectum is \( |4p| \).
Length of latus rectum \( = \frac{1}{2} \).
In simple words: We found the parabola's equation by making it pass through three given points, leading to \( y = 2x^2 + 3x + 4 \). Then, we rearranged this equation into a special form \( (x + \frac{3}{4})^2 = \frac{1}{2}(y - \frac{23}{8}) \). From this form, we could directly see that the length of its latus rectum (a measure of its width) is \( \frac{1}{2} \).

🎯 Exam Tip: The latus rectum is directly obtained from the coefficient of the linear term when the equation is in standard form. For \( (x-h)^2 = 4p(y-k) \), the latus rectum is \( 4|p| \). Remember to factor out the coefficient of the linear term correctly before identifying \( 4p \).

Question 18. Find the coordinates of the point on the parabola y² = 8x whose focal distance is 8.
Answer: The given equation of the parabola is \( y^2 = 8x \).
We compare this equation to the standard form \( y^2 = 4ax \).
From \( 4a = 8 \), we find \( a = \frac{8}{4} = 2 \).
The focus of this parabola is at \( (a, 0) \), which is \( (2, 0) \).
The focal distance of any point \( (x_1, y_1) \) on the parabola \( y^2 = 4ax \) is defined as \( x_1 + a \).
We are given that the focal distance is 8.
So, \( x_1 + a = 8 \).
Substitute the value of \( a = 2 \):
\( x_1 + 2 = 8 \).
\( x_1 = 8 - 2 = 6 \).
Now that we have the x-coordinate \( x_1 = 6 \), we substitute it back into the parabola's equation \( y^2 = 8x \) to find \( y_1 \).
\( y_1^2 = 8(6) \)
\( y_1^2 = 48 \).
Take the square root of both sides:
\( y_1 = \pm\sqrt{48} \).
To simplify \( \sqrt{48} \), we find the largest perfect square factor: \( \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \).
So, \( y_1 = \pm 4\sqrt{3} \).
Therefore, the coordinates of the points on the parabola whose focal distance is 8 are \( (6, 4\sqrt{3}) \) and \( (6, -4\sqrt{3}) \). These points are equidistant from the focus and the directrix.
In simple words: For the parabola \( y^2 = 8x \), we found its special value \( a=2 \). The focal distance (distance from a point on the parabola to the focus) is given as 8. Using the formula for focal distance, \( x_1 + a = 8 \), we found the x-coordinate of the point is 6. Plugging \( x=6 \) back into the parabola's equation, we found the y-coordinates to be \( \pm 4\sqrt{3} \).

🎯 Exam Tip: Remember the definition of focal distance: for a parabola \( y^2 = 4ax \), the focal distance of a point \( (x_1, y_1) \) is \( x_1 + a \). For \( x^2 = 4ay \), it's \( y_1 + a \). Use this relationship to find the unknown coordinate when one is given.

Question 19. If the ordinate of a point on the parabola y² = 4ax is twice the latus rectum, prove that the abscissa of this point is twice the ordinate.
Answer: Let the point on the parabola \( y^2 = 4ax \) be \( (x_1, y_1) \).
From the equation of the parabola, we know that the length of the latus rectum is \( 4a \).
We are given that the ordinate \( (y_1) \) of this point is twice the latus rectum.
So, \( y_1 = 2 \times (\text{length of latus rectum}) \).
\( y_1 = 2 \times (4a) \).
\( y_1 = 8a \).

Now, since the point \( (x_1, y_1) \) lies on the parabola \( y^2 = 4ax \), it must satisfy the equation:
\( y_1^2 = 4ax_1 \).
Substitute \( y_1 = 8a \) into this equation:
\( (8a)^2 = 4ax_1 \).
\( 64a^2 = 4ax_1 \).
To find \( x_1 \), divide both sides by \( 4a \) (assuming \( a \ne 0 \), which it must be for a parabola).
\( x_1 = \frac{64a^2}{4a} \).
\( x_1 = 16a \).

We need to prove that the abscissa \( (x_1) \) is twice the ordinate \( (y_1) \).
We found \( x_1 = 16a \) and \( y_1 = 8a \).
Let's check if \( x_1 = 2y_1 \):
\( 2y_1 = 2(8a) = 16a \).
Since \( x_1 = 16a \) and \( 2y_1 = 16a \), it is proven that \( x_1 = 2y_1 \).
Therefore, the abscissa of the point is twice its ordinate. This shows a special relationship between the coordinates for points meeting this condition.
In simple words: We have a point on the parabola \( y^2 = 4ax \). We are told its 'y' value (ordinate) is twice the latus rectum, which is \( 8a \). By putting this \( y \) value into the parabola's equation, we find the 'x' value (abscissa) to be \( 16a \). Since \( 16a \) is twice \( 8a \), we proved that the 'x' value is twice the 'y' value.

🎯 Exam Tip: Problems that ask to "prove" a relationship often involve substituting given conditions into the fundamental equations and simplifying them algebraically. Ensure your steps are clear and logical to demonstrate the desired relationship.

Question 20. Find the equation of the parabola whose focus is at the origin, and whose directrix is the line y – x = 4. Find also the length of the latus rectum, the equation of the axis, and the coordinates of the vertex.
Answer: Given that the focus \( F \) is at the origin \( (0, 0) \) and the directrix is the line \( y - x = 4 \), which can be written as \( -x + y - 4 = 0 \).

**1. Equation of the parabola:**
Let \( P(x, y) \) be any point on the parabola. By definition, \( |PF| = |PM| \) (distance from point to focus equals distance from point to directrix).
\( \sqrt{(x - 0)^2 + (y - 0)^2} = \frac{|-x + y - 4|}{\sqrt{(-1)^2 + (1)^2}} \).
\( \sqrt{x^2 + y^2} = \frac{|-x + y - 4|}{\sqrt{1 + 1}} \).
\( \sqrt{x^2 + y^2} = \frac{|y - x - 4|}{\sqrt{2}} \).
Square both sides:
\( x^2 + y^2 = \frac{(y - x - 4)^2}{2} \).
\( 2(x^2 + y^2) = (y - x - 4)^2 \).
\( 2x^2 + 2y^2 = y^2 + x^2 + 16 - 2xy - 8y + 8x \).
Rearrange terms to get the equation of the parabola:
\( 2x^2 - x^2 + 2y^2 - y^2 + 2xy - 8x + 8y - 16 = 0 \).
\( x^2 + y^2 + 2xy - 8x + 8y - 16 = 0 \). This is the required equation of the parabola.

**2. Length of the latus rectum:**
The length of the latus rectum for a parabola is \( 2 \times (\text{distance from focus to directrix}) \).
Distance from focus \( (0,0) \) to directrix \( -x + y - 4 = 0 \) is:
\( d = \frac{|-(0) + (0) - 4|}{\sqrt{(-1)^2 + (1)^2}} = \frac{|-4|}{\sqrt{2}} = \frac{4}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2} \).
Length of latus rectum \( = 2d = 2(2\sqrt{2}) = 4\sqrt{2} \).

**3. Equation of the axis:**
The axis of the parabola is the line passing through the focus \( (0,0) \) and perpendicular to the directrix \( -x + y - 4 = 0 \).
The slope of the directrix \( (y = x + 4) \) is \( m_d = 1 \).
The slope of the axis \( m_a \) must be \( -1 \) (since \( m_d \times m_a = -1 \)).
The equation of the axis passing through \( (0,0) \) with slope \( -1 \) is \( y - 0 = -1(x - 0) \).
\( y = -x \), or \( x + y = 0 \). This is the required equation of the axis.

**4. Coordinates of the vertex:**
The vertex \( (h, k) \) is the midpoint of the focus \( F(0,0) \) and the point where the axis intersects the directrix. Let's find this intersection point \( Z \).
Solve the directrix \( y - x = 4 \) and the axis \( y = -x \) simultaneously.
Substitute \( y = -x \) into \( y - x = 4 \):
\( (-x) - x = 4 \)
\( -2x = 4 \)
\( x = -2 \).
Then, \( y = -(-2) = 2 \).
So, the point of intersection \( Z \) is \( (-2, 2) \).

The vertex \( (h, k) \) is the midpoint of \( F(0,0) \) and \( Z(-2, 2) \).
\( h = \frac{0 + (-2)}{2} = \frac{-2}{2} = -1 \).
\( k = \frac{0 + 2}{2} = \frac{2}{2} = 1 \).
The coordinates of the vertex are \( (-1, 1) \).
In simple words: The parabola has its focus at \( (0,0) \) and its directrix is \( y-x=4 \). Using the definition of a parabola, we found its equation to be \( x^2 + y^2 + 2xy - 8x + 8y - 16 = 0 \). The length of its latus rectum is \( 4\sqrt{2} \). The axis of the parabola is the line \( x+y=0 \). Finally, its vertex (the midpoint between the focus and the directrix) is at \( (-1, 1) \).

🎯 Exam Tip: For complex parabolas (with tilted axes), remember the full set of properties: equation from \( PF=PM \), latus rectum from \( 2 \times \) (focus-directrix distance), axis as a line through the focus perpendicular to the directrix, and vertex as the midpoint of the focus and the directrix-axis intersection point. Each step must be calculated accurately.

Question 21. The directrix of a conic section is the straight line 3x – 4y + 5 = 0 and the focus is (2, 3). If the eccentricity e is 1, find the equation to the conic section. Is the conic section a parabola ?
Answer: Given the directrix is the line \( 3x - 4y + 5 = 0 \), and the focus \( F \) is at \( (2, 3) \). The eccentricity \( e \) is given as 1.
A conic section is a parabola if and only if its eccentricity \( e = 1 \). Since \( e = 1 \), the given conic section is indeed a parabola.
To find the equation of the parabola, let \( P(x, y) \) be any point on the parabola. By the definition of a conic section, the ratio of the distance from \( P \) to the focus \( F \) to the perpendicular distance from \( P \) to the directrix \( PM \) is equal to the eccentricity \( e \). So, \( |PF| = e|PM| \). Since \( e=1 \), we have \( |PF| = |PM| \).
\( \sqrt{(x - 2)^2 + (y - 3)^2} = \frac{|3x - 4y + 5|}{\sqrt{3^2 + (-4)^2}} \).
\( \sqrt{(x - 2)^2 + (y - 3)^2} = \frac{|3x - 4y + 5|}{\sqrt{9 + 16}} \).
\( \sqrt{(x - 2)^2 + (y - 3)^2} = \frac{|3x - 4y + 5|}{\sqrt{25}} \).
\( \sqrt{(x - 2)^2 + (y - 3)^2} = \frac{|3x - 4y + 5|}{5} \).
Square both sides:
\( (x - 2)^2 + (y - 3)^2 = \frac{(3x - 4y + 5)^2}{25} \).
Multiply both sides by 25:
\( 25[(x - 2)^2 + (y - 3)^2] = (3x - 4y + 5)^2 \).
Expand the terms:
\( 25[(x^2 - 4x + 4) + (y^2 - 6y + 9)] = (9x^2 + 16y^2 + 25 - 24xy + 30x - 40y) \).
\( 25[x^2 + y^2 - 4x - 6y + 13] = 9x^2 + 16y^2 + 25 - 24xy + 30x - 40y \).
\( 25x^2 + 25y^2 - 100x - 150y + 325 = 9x^2 + 16y^2 - 24xy + 30x - 40y + 25 \).
Move all terms to the left side:
\( (25x^2 - 9x^2) + (25y^2 - 16y^2) + 24xy - 100x - 30x - 150y + 40y + 325 - 25 = 0 \).
\( 16x^2 + 9y^2 + 24xy - 130x - 110y + 300 = 0 \). This is the required equation of the parabola.
Yes, the conic section is a parabola because its eccentricity \( e = 1 \). This confirms the type of conic section.
In simple words: The problem gives a directrix line and a focus point for a shape with eccentricity \( e=1 \). Because \( e=1 \), we know for sure it's a parabola. We used the rule that any point on the parabola is equally distant from the focus and the directrix. After doing some algebra, we found the equation of this parabola to be \( 16x^2 + 9y^2 + 24xy - 130x - 110y + 300 = 0 \).

🎯 Exam Tip: The eccentricity \( e \) is the defining characteristic of a conic section: \( e=1 \) for a parabola, \( e<1 \) for an ellipse, and \( e>1 \) for a hyperbola. Always state this explicitly when \( e \) is given.

Question 22. Find the equation to the parabola whose focus is (-2, 1) and directrix is 6x – 3y = 8.
Answer: Given that the focus \( F \) is at \( (-2, 1) \) and the directrix is the line \( 6x - 3y = 8 \), which can be written as \( 6x - 3y - 8 = 0 \).
Let \( P(x, y) \) be any point on the parabola. By the definition of a parabola, the distance from \( P \) to the focus \( F \) is equal to the perpendicular distance from \( P \) to the directrix \( PM \). So, \( |PF| = |PM| \).
\( \sqrt{(x - (-2))^2 + (y - 1)^2} = \frac{|6x - 3y - 8|}{\sqrt{6^2 + (-3)^2}} \).
\( \sqrt{(x + 2)^2 + (y - 1)^2} = \frac{|6x - 3y - 8|}{\sqrt{36 + 9}} \).
\( \sqrt{(x + 2)^2 + (y - 1)^2} = \frac{|6x - 3y - 8|}{\sqrt{45}} \).
Since \( \sqrt{45} = \sqrt{9 \times 5} = 3\sqrt{5} \), we have:
\( \sqrt{(x + 2)^2 + (y - 1)^2} = \frac{|6x - 3y - 8|}{3\sqrt{5}} \).
Square both sides:
\( (x + 2)^2 + (y - 1)^2 = \frac{(6x - 3y - 8)^2}{(3\sqrt{5})^2} \).
\( (x + 2)^2 + (y - 1)^2 = \frac{(6x - 3y - 8)^2}{45} \).
Multiply both sides by 45:
\( 45[(x + 2)^2 + (y - 1)^2] = (6x - 3y - 8)^2 \).
Expand the terms:
\( 45[(x^2 + 4x + 4) + (y^2 - 2y + 1)] = (36x^2 + 9y^2 + 64 - 36xy - 96x + 48y) \).
\( 45[x^2 + y^2 + 4x - 2y + 5] = 36x^2 + 9y^2 + 64 - 36xy - 96x + 48y \).
\( 45x^2 + 45y^2 + 180x - 90y + 225 = 36x^2 + 9y^2 - 36xy - 96x + 48y + 64 \).
Move all terms to one side:
\( (45x^2 - 36x^2) + (45y^2 - 9y^2) + 36xy + 180x + 96x - 90y - 48y + 225 - 64 = 0 \).
\( 9x^2 + 36y^2 + 36xy + 276x - 138y + 161 = 0 \). This is the required equation of the parabola.
In simple words: To find the parabola's equation, we set the distance from any point \( (x,y) \) to the focus \( (-2,1) \) equal to its perpendicular distance to the directrix \( 6x-3y=8 \). After squaring both sides and simplifying the expression, we found the equation of the parabola is \( 9x^2 + 36y^2 + 36xy + 276x - 138y + 161 = 0 \).

🎯 Exam Tip: Carefully handle the calculations, especially when squaring trinomials like \( (6x - 3y - 8)^2 \). It's easy to miss terms or make sign errors. Double-check each expansion step to maintain accuracy.

Question 23. Find the length of the latus rectum of the parabola whose focus is (3, 3) and directrix is 3x – 4y - 2 = 0 is
(a) 2
(b) 1
(c) 4
(d) None of the options
Answer: (a) 2
The length of the latus rectum of a parabola is equal to twice the perpendicular distance from the focus to the directrix. Let \( F(x_1, y_1) \) be the focus and \( Ax + By + C = 0 \) be the equation of the directrix. The perpendicular distance \( d \) is given by the formula: \( d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \).
Given focus \( (3, 3) \) and directrix \( 3x - 4y - 2 = 0 \).
Here, \( x_1 = 3 \), \( y_1 = 3 \), \( A = 3 \), \( B = -4 \), \( C = -2 \).
Calculate the perpendicular distance \( d \):
\( d = \frac{|3(3) - 4(3) - 2|}{\sqrt{3^2 + (-4)^2}} \).
\( d = \frac{|9 - 12 - 2|}{\sqrt{9 + 16}} \).
\( d = \frac{|-5|}{\sqrt{25}} \).
\( d = \frac{5}{5} = 1 \).
The length of the latus rectum is \( 2d \).
Length of latus rectum \( = 2 \times 1 = 2 \).
Therefore, the correct option is (a). The distance from the focus to the directrix gives a direct measure related to the parabola's width.
In simple words: The latus rectum is twice the distance from the focus to the directrix. We calculated this distance using the given focus \( (3,3) \) and directrix \( 3x-4y-2=0 \). The distance was 1, so the latus rectum is \( 2 \times 1 = 2 \).

🎯 Exam Tip: Remember the direct formula for the length of the latus rectum for a parabola when the focus and directrix are known. It is always \( 2 \times (\text{perpendicular distance from focus to directrix}) \).

Question 24. The length of the latus rectum of the parabola whose focus is (3, 3) and directrix is 3x - 4y - 2 = 0 is
(a) 2
(b) 1
(c) 4
(d) None of the options
Answer: (a) 2
Required length of latus rectum is found by taking twice the perpendicular distance from the focus to the directrix. We use the formula for the perpendicular distance from a point \( (x_1, y_1) \) to a line \( Ax + By + C = 0 \), which is \( \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \). This length is an important property that describes how wide the parabola is at its focus.
\( \text{Length of latus rectum} = 2 \times \frac{|3(3) - 4(3) - 2|}{\sqrt{3^2 + (-4)^2}} \)
\( = 2 \times \frac{|9 - 12 - 2|}{\sqrt{9 + 16}} \)
\( = 2 \times \frac{|-5|}{\sqrt{25}} \)
\( = 2 \times \frac{5}{5} \)
\( = 2 \times 1 \)
\( = 2 \)
In simple words: To find the length of the latus rectum, we calculate the shortest distance from the focus point (3,3) to the directrix line \( 3x - 4y - 2 = 0 \), and then we multiply that distance by 2. After calculating, the distance is 1, so the total length of the latus rectum is 2.

🎯 Exam Tip: Remember the formula that the length of the latus rectum for a parabola is twice the perpendicular distance from its focus to its directrix. This is a common question type.