OP Malhotra Class 11 Maths Solutions Chapter 23 Parabola Chapter Test

Question 1. The equation of the directrix of the parabola is 3x + 2y + 1 = 0. The focus is (2, 1). Find the equation of the parabola.
Answer: Let the focus of the parabola be \(F(2,1)\) and the equation of the directrix be \(3x + 2y + 1 = 0\). If \(P(x,y)\) is any point on the parabola, the distance from P to the focus (PF) must be equal to the perpendicular distance from P to the directrix (PM). This is the definition of a parabola.
So, \(PF = PM\)
\( \implies \sqrt{(x - 2)^2 + (y - 1)^2} = \frac{|3x + 2y + 1|}{\sqrt{3^2 + 2^2}} \)
\( \implies \sqrt{(x - 2)^2 + (y - 1)^2} = \frac{|3x + 2y + 1|}{\sqrt{9 + 4}} \)
\( \implies \sqrt{(x - 2)^2 + (y - 1)^2} = \frac{|3x + 2y + 1|}{\sqrt{13}} \)
Now, we square both sides of the equation to remove the square root:
\( \implies (x - 2)^2 + (y - 1)^2 = \frac{(3x + 2y + 1)^2}{13} \)
\( \implies 13 [(x^2 - 4x + 4) + (y^2 - 2y + 1)] = (3x + 2y + 1)^2 \)
\( \implies 13 [x^2 - 4x + y^2 - 2y + 5] = (9x^2 + 4y^2 + 1 + 12xy + 4y + 6x) \)
\( \implies 13x^2 - 52x + 13y^2 - 26y + 65 = 9x^2 + 4y^2 + 1 + 12xy + 4y + 6x \)
Rearranging all terms to one side gives the equation of the parabola:
\( \implies 13x^2 - 9x^2 + 13y^2 - 4y^2 - 12xy - 52x - 6x - 26y - 4y + 65 - 1 = 0 \)
\( \implies 4x^2 + 9y^2 - 12xy - 58x - 30y + 64 = 0 \)
In simple words: To find the parabola's equation, we use the rule that any point on the parabola is the same distance from the special "focus" point and the special "directrix" line. We write this as an equation, then square both sides to simplify and get the final equation.

🎯 Exam Tip: Remember to always square both sides of the PF=PM equation to eliminate the square roots, and then expand carefully. A common mistake is forgetting to square the denominator as well.

Question 2. The points (0, 4) and (0, 2) are the vertex and focus of a parabola. Find the equation of the parabola.
Answer: We are given the vertex \(V(0,4)\) and the focus \(F(0,2)\). Since the x-coordinates of both the vertex and focus are the same (both are 0), the axis of the parabola is the y-axis. The vertex is at \( (0,4) \) and the focus is at \( (0,2) \). This means the vertex is above the focus, so the parabola opens downwards. The distance between the vertex and the focus is \( a \).
\( a = \text{distance from V to F} = \sqrt{(0-0)^2 + (4-2)^2} = \sqrt{0^2 + 2^2} = \sqrt{4} = 2 \).
So, \( a = 2 \).
For a downward-opening parabola with vertex at \((h,k)\), the equation is \( (x-h)^2 = -4a(y-k) \).
Here, \((h,k) = (0,4)\) and \(a=2\).
Substituting these values:
\( (x-0)^2 = -4(2)(y-4) \)
\( \implies x^2 = -8(y-4) \)
\( \implies x^2 = -8y + 32 \)
Rearranging the terms to get the standard form:
\( \implies x^2 + 8y - 32 = 0 \)
This is the required equation of the parabola. The latus rectum length is \( 4a = 4 \times 2 = 8 \).

In simple words: When the special vertex point is higher than the special focus point and they are both on the y-axis, the parabola will open downwards. We calculate the distance between these two points to find 'a', then use a simple formula for downward parabolas to get the final equation.

🎯 Exam Tip: Always sketch a quick diagram to understand the orientation of the parabola (opens up, down, left, or right) based on the vertex and focus positions. This helps choose the correct standard form of the equation.

Question 3. Find the equation of the parabola with latus rectum joining points (4, 6) and (4,-2).
Answer: We are given the endpoints of the latus rectum as \(L(4,6)\) and \(L'(4,-2)\).
Since the x-coordinates of both points are the same (both are 4), the latus rectum is a vertical line. This means the axis of the parabola is perpendicular to the latus rectum, so the axis must be parallel to the x-axis. Therefore, the general equation of the parabola will be of the form \( (y-k)^2 = \pm 4a(x-h) \), where \((h,k)\) is the vertex.
The length of the latus rectum is the distance between \(L\) and \(L'\):
\( \text{Length of latus rectum} = |LL'| = \sqrt{(4-4)^2 + (6-(-2))^2} \)
\( \implies = \sqrt{0^2 + (6+2)^2} = \sqrt{8^2} = 8 \).
We know that the length of the latus rectum is \(4a\). So, \(4a = 8 \implies a = 2\).
The midpoint of the latus rectum is the focus of the parabola. The midpoint is \( \left( \frac{4+4}{2}, \frac{6+(-2)}{2} \right) = \left( \frac{8}{2}, \frac{4}{2} \right) = (4,2) \). So, the focus is \(F(4,2)\).
Since the axis is parallel to the x-axis, the y-coordinate of the vertex will be the same as the y-coordinate of the focus, i.e., \(k=2\).
The vertex \((h,k)\) is at a distance 'a' from the focus along the axis. Since the latus rectum is at \(x=4\), the parabola can open either to the right or to the left from \(x=4\).
Case 1: Parabola opens to the right.
The vertex \(A(h,k)\) is \(a\) units to the left of the focus. So, \(h = 4 - a = 4 - 2 = 2\).
The vertex is \(A(2,2)\).
The equation of the parabola is \( (y-k)^2 = 4a(x-h) \)
\( \implies (y-2)^2 = 4(2)(x-2) \)
\( \implies (y-2)^2 = 8(x-2) \)
\( \implies y^2 - 4y + 4 = 8x - 16 \)
\( \implies y^2 - 4y - 8x + 20 = 0 \).
Case 2: Parabola opens to the left.
The vertex \(A'(h',k)\) is \(a\) units to the right of the focus. So, \(h' = 4 + a = 4 + 2 = 6\).
The vertex is \(A'(6,2)\).
The equation of the parabola is \( (y-k)^2 = -4a(x-h') \)
\( \implies (y-2)^2 = -4(2)(x-6) \)
\( \implies (y-2)^2 = -8(x-6) \)
\( \implies y^2 - 4y + 4 = -8x + 48 \)
\( \implies y^2 - 4y + 8x - 44 = 0 \).
Thus, there are two possible equations for the parabola depending on its direction.

In simple words: When you know the two end points of the latus rectum, you can find its length and the middle point, which is the focus. Since the latus rectum is vertical, the parabola opens sideways (either left or right). Depending on which way it opens, you'll find two possible vertex points, and each will give you a different equation for the parabola.

🎯 Exam Tip: When given the endpoints of the latus rectum, always find its length (to get 'a') and its midpoint (which is the focus). Remember to consider both possible orientations of the parabola (e.g., opening left or right if the latus rectum is vertical) to find all possible equations.

Question 4. Find the equation of the parabola whose focus is (- 1, – 2) and the equation of the directrix is given 4x – 3y + 2 = 0. Also find the equation of the axis.
Answer: We are given the focus \(F(-1,-2)\) and the equation of the directrix \(4x - 3y + 2 = 0\).
Let \(P(x,y)\) be any point on the parabola. By definition, the distance from P to the focus (PF) is equal to the perpendicular distance from P to the directrix (PM).
\(PF = PM\)
\( \implies \sqrt{(x - (-1))^2 + (y - (-2))^2} = \frac{|4x - 3y + 2|}{\sqrt{4^2 + (-3)^2}} \)
\( \implies \sqrt{(x + 1)^2 + (y + 2)^2} = \frac{|4x - 3y + 2|}{\sqrt{16 + 9}} \)
\( \implies \sqrt{(x + 1)^2 + (y + 2)^2} = \frac{|4x - 3y + 2|}{\sqrt{25}} \)
\( \implies \sqrt{(x + 1)^2 + (y + 2)^2} = \frac{|4x - 3y + 2|}{5} \)
Square both sides to remove the square root:
\( \implies (x + 1)^2 + (y + 2)^2 = \frac{(4x - 3y + 2)^2}{25} \)
\( \implies 25 [(x^2 + 2x + 1) + (y^2 + 4y + 4)] = (16x^2 + 9y^2 + 4 - 24xy - 12y + 16x) \)
\( \implies 25 [x^2 + y^2 + 2x + 4y + 5] = 16x^2 + 9y^2 - 24xy + 16x - 12y + 4 \)
\( \implies 25x^2 + 25y^2 + 50x + 100y + 125 = 16x^2 + 9y^2 - 24xy + 16x - 12y + 4 \)
Rearranging all terms to one side gives the equation of the parabola:
\( \implies (25-16)x^2 + (25-9)y^2 + 24xy + (50-16)x + (100+12)y + (125-4) = 0 \)
\( \implies 9x^2 + 16y^2 + 24xy + 34x + 112y + 121 = 0 \)
This is the required equation of the parabola.

Now, let's find the equation of the axis.
The axis of the parabola is a line that passes through the focus \(F(-1,-2)\) and is perpendicular to the directrix \(4x - 3y + 2 = 0\).
The slope of the directrix is \( m_d = - \frac{\text{coefficient of x}}{\text{coefficient of y}} = - \frac{4}{-3} = \frac{4}{3} \).
The slope of the axis \(m_a\) will be the negative reciprocal of \(m_d\) because they are perpendicular:
\( m_a = - \frac{1}{m_d} = - \frac{1}{4/3} = - \frac{3}{4} \).
Using the point-slope form \( y - y_1 = m(x - x_1) \) with focus \(F(-1,-2)\) and slope \(m_a = - \frac{3}{4}\):
\( y - (-2) = - \frac{3}{4}(x - (-1)) \)
\( \implies y + 2 = - \frac{3}{4}(x + 1) \)
\( \implies 4(y + 2) = -3(x + 1) \)
\( \implies 4y + 8 = -3x - 3 \)
\( \implies 3x + 4y + 8 + 3 = 0 \)
\( \implies 3x + 4y + 11 = 0 \)
This is the required equation of the axis of the parabola.
In simple words: To find the parabola's equation, we use the rule that any point on the parabola is equally far from the special focus point and the special directrix line. We set these distances equal, then square and simplify to get the final equation. For the axis, we find a line that goes through the focus and is exactly perpendicular to the directrix line, using slopes and point-slope form.

🎯 Exam Tip: When finding the equation of the axis, remember it's always perpendicular to the directrix and passes through the focus. Use the relationship between slopes of perpendicular lines to find the axis's slope, then apply the point-slope formula.

Question 5. Find the equation of the parabola if its vertex is at (0, 0), passes through (5, 2) and is symmetric w.r.t. y-axis.
Answer: We are given that the vertex of the parabola is at the origin \( (0,0) \) and it is symmetric with respect to the y-axis. This means the standard form of the parabola's equation is \( x^2 = 4ay \).
The parabola also passes through the point \( (5,2) \). We can substitute these coordinates into the equation to find the value of \( a \).
Substitute \( x=5 \) and \( y=2 \) into \( x^2 = 4ay \):
\( (5)^2 = 4a(2) \)
\( \implies 25 = 8a \)
To find \( a \), divide both sides by 8:
\( \implies a = \frac{25}{8} \)
Now, substitute the value of \( a \) back into the standard equation \( x^2 = 4ay \):
\( x^2 = 4 \left( \frac{25}{8} \right) y \)
Simplify the right side:
\( x^2 = \frac{100}{8} y \)
\( \implies x^2 = \frac{25}{2} y \)
This is the required equation of the parabola.
In simple words: We know the parabola starts at (0,0) and is balanced around the y-axis, so its basic shape is \( x^2 = 4ay \). Since it goes through the point (5,2), we put these numbers into the equation to find the missing value 'a'. Once we have 'a', we put it back into the basic shape to get the final equation.

🎯 Exam Tip: When a parabola is symmetric about an axis and has its vertex at the origin, immediately recall the two simplest standard forms: \( y^2 = 4ax \) (symmetric about x-axis) or \( x^2 = 4ay \) (symmetric about y-axis). Using the given point helps determine 'a' and thus the full equation.

Question 6. Find the length of its latus rectum.
Answer: Let the given equation of the parabola be in the standard form \( y^2 = 4ax \).
We are told that this parabola passes through the point \( (2,-6) \). We can use this point to find the value of \( a \).
Substitute \( x=2 \) and \( y=-6 \) into the equation \( y^2 = 4ax \):
\( (-6)^2 = 4a(2) \)
\( \implies 36 = 8a \)
To find \( a \), divide both sides by 8:
\( \implies a = \frac{36}{8} \)
Simplify the fraction for \( a \):
\( \implies a = \frac{9}{2} \)
The length of the latus rectum for a parabola in the form \( y^2 = 4ax \) is given by \( 4a \).
Now, substitute the value of \( a = \frac{9}{2} \) into the latus rectum formula:
\( \text{Length of latus rectum} = 4a = 4 \times \frac{9}{2} \)
\( \implies = \frac{36}{2} \)
\( \implies = 18 \)
So, the length of the latus rectum is 18 units.
In simple words: For a simple parabola like \( y^2 = 4ax \), the length of its "latus rectum" is \( 4a \). We use the point it passes through to find what 'a' is. Once we know 'a', we just multiply it by 4 to get the length.

🎯 Exam Tip: The length of the latus rectum is always \( |4a| \). Make sure to correctly identify the standard form of the parabola from the given information to find the parameter 'a' accurately.

Question 7. Find the coordinates of the vertex and the focus of the parabola \( y^2 = 4(x + y) \).
Answer: We are given the equation of the parabola \( y^2 = 4(x + y) \).
To find the vertex and focus, we need to rewrite this equation into a standard form like \( (y-k)^2 = 4a(x-h) \) or \( (x-h)^2 = 4a(y-k) \).
First, expand the equation and move all y-terms to one side and x-terms to the other:
\( y^2 = 4x + 4y \)
\( \implies y^2 - 4y = 4x \)
Now, complete the square for the y-terms. To do this, we add \( \left(\frac{\text{coefficient of y}}{2}\right)^2 \) to both sides. The coefficient of y is -4, so we add \( \left(\frac{-4}{2}\right)^2 = (-2)^2 = 4 \).
\( y^2 - 4y + 4 = 4x + 4 \)
This simplifies to:
\( \implies (y - 2)^2 = 4(x + 1) \)
This equation is in the standard form \( Y^2 = 4aX \), where \( Y = y - 2 \) and \( X = x + 1 \).
Comparing \( (y-2)^2 = 4(x+1) \) with \( Y^2 = 4aX \), we can see that:
\( Y = y - 2 \)
\( X = x + 1 \)
\( 4a = 4 \implies a = 1 \).
This represents a right-handed parabola (opens to the right) in the new coordinate system \( (X,Y) \).

**1. Finding the Vertex:**
The vertex of the standard parabola \( Y^2 = 4aX \) is \( (0,0) \) in the \( (X,Y) \) system.
So, we set \( X=0 \) and \( Y=0 \):
\( x + 1 = 0 \implies x = -1 \)
\( y - 2 = 0 \implies y = 2 \)
Thus, the coordinates of the vertex of the given parabola are \( (-1, 2) \).

**2. Finding the Focus:**
The focus of the standard parabola \( Y^2 = 4aX \) is \( (a,0) \) in the \( (X,Y) \) system.
Since \( a=1 \), the focus in the \( (X,Y) \) system is \( (1,0) \).
Now, convert back to \( (x,y) \) coordinates:
\( X = 1 \implies x + 1 = 1 \implies x = 0 \)
\( Y = 0 \implies y - 2 = 0 \implies y = 2 \)
Thus, the coordinates of the focus of the given parabola are \( (0, 2) \).
In simple words: We start by rearranging the given equation to match a common parabola form by completing the square. This helps us find 'a' and identify new 'X' and 'Y' values. The vertex is where X and Y are zero, and the focus is at 'a' distance from the vertex along the axis. We then convert these back to the original 'x' and 'y' coordinates.

🎯 Exam Tip: Always complete the square for the squared term (y-term in this case) to transform the equation into a recognizable standard form. This makes it easier to identify the vertex, focus, and directrix relative to the shifted origin.

Question 8. Find the focus, the equation of the directrix and the length of the latus rectum of the parabola \( y^2 - 4x - 4y + 12 = 0 \).
Answer: We are given the equation of the parabola \( y^2 - 4x - 4y + 12 = 0 \).
To find its focus, directrix, and latus rectum, we need to rewrite this equation in a standard form like \( (y-k)^2 = 4a(x-h) \).
First, move all terms involving \( x \) and constants to the right side:
\( y^2 - 4y = 4x - 12 \)
Now, complete the square for the y-terms on the left side. The coefficient of y is -4, so we add \( \left(\frac{-4}{2}\right)^2 = (-2)^2 = 4 \) to both sides:
\( y^2 - 4y + 4 = 4x - 12 + 4 \)
This simplifies to:
\( \implies (y - 2)^2 = 4x - 8 \)
Factor out 4 from the right side:
\( \implies (y - 2)^2 = 4(x - 2) \)
This equation is in the standard form \( Y^2 = 4aX \), where \( Y = y - 2 \) and \( X = x - 2 \).
Comparing \( (y-2)^2 = 4(x-2) \) with \( Y^2 = 4aX \), we can identify:
\( Y = y - 2 \)
\( X = x - 2 \)
\( 4a = 4 \implies a = 1 \).
This represents a right-handed parabola in the new coordinate system \( (X,Y) \). The vertex is at \( (h,k) = (2,2) \).

**1. Length of the Latus Rectum:**
The length of the latus rectum for a parabola of the form \( Y^2 = 4aX \) is \( 4a \).
Since \( a=1 \), the length of the latus rectum is \( 4(1) = 4 \) units.

**2. Focus of the Parabola:**
The focus of the standard parabola \( Y^2 = 4aX \) is \( (a,0) \) in the \( (X,Y) \) system.
Since \( a=1 \), the focus in the \( (X,Y) \) system is \( (1,0) \).
Now, convert back to \( (x,y) \) coordinates:
\( X = 1 \implies x - 2 = 1 \implies x = 3 \)
\( Y = 0 \implies y - 2 = 0 \implies y = 2 \)
Thus, the coordinates of the focus of the given parabola are \( (3, 2) \).

**3. Equation of the Directrix:**
The equation of the directrix for the standard parabola \( Y^2 = 4aX \) is \( X = -a \) in the \( (X,Y) \) system.
Since \( a=1 \), the directrix in the \( (X,Y) \) system is \( X = -1 \).
Now, convert back to \( (x,y) \) coordinates:
\( X = -1 \implies x - 2 = -1 \implies x = 1 \)
Thus, the equation of the directrix is \( x = 1 \).
In simple words: We first change the parabola's equation into a simpler form by moving terms and completing the square. This helps us find 'a', which is key to everything else. The length of the latus rectum is simply \( 4a \). The focus is found by adding 'a' to the x-coordinate of the vertex, and the directrix is a line found by subtracting 'a' from the x-coordinate of the vertex.

🎯 Exam Tip: After converting the equation to standard form \( (y-k)^2 = 4a(x-h) \) or \( (x-h)^2 = 4a(y-k) \), clearly identify \( h, k, \) and \( a \). Then, use the standard formulas for focus, directrix, and latus rectum, remembering to translate back to the original \( (x,y) \) coordinates.