RBSE Solutions Class 11 Maths Chapter 12 Conic Section Exercise 12.3

Question 1. Find the equation of parabola whose
(i) Focus (2, 3) and directrix x – 4y + 3 = 0
(ii) Focus (-3, 0) and directri x + 5 = 0
Answer:
(i) Let \(P(h, k)\) be any point on the parabola. According to the definition of a parabola, the distance from \(P\) to the focus (SP) is equal to the perpendicular distance from \(P\) to the directrix (PM).
So, \( SP = PM \)
Squaring both sides, we get \( SP^2 = PM^2 \)
The distance formula for SP and the perpendicular distance formula for PM give:
\( (h - 2)^2 + (k - 3)^2 = \left( \frac{h - 4k + 3}{\sqrt{1^2 + (-4)^2}} \right)^2 \)
\( (h - 2)^2 + (k - 3)^2 = \frac{(h - 4k + 3)^2}{1 + 16} \)
\( (h^2 - 4h + 4) + (k^2 - 6k + 9) = \frac{h^2 + 16k^2 + 9 - 8hk + 6h - 24k}{17} \)
Now, multiply both sides by 17:
\( 17(h^2 - 4h + 4 + k^2 - 6k + 9) = h^2 + 16k^2 + 9 - 8hk + 6h - 24k \)
\( 17h^2 - 68h + 68 + 17k^2 - 102k + 153 = h^2 + 16k^2 + 9 - 8hk + 6h - 24k \)
Move all terms to one side:
\( 17h^2 - h^2 + 17k^2 - 16k^2 - 68h - 6h - 102k + 24k + 68 + 153 - 9 + 8hk = 0 \)
\( 16h^2 + k^2 - 74h - 78k + 212 + 8hk = 0 \)
So, the locus of point \( P(h, k) \) is \( 16x^2 + y^2 - 74x - 78y + 8xy + 212 = 0 \).
(ii) Let \(P(h, k)\) be any point on the parabola. The focus is \( (-3, 0) \) and the directrix is \( x + 5 = 0 \).
Using the definition \( SP^2 = PM^2 \):
\( (h - (-3))^2 + (k - 0)^2 = \left( \frac{h + 5}{\sqrt{1^2 + 0^2}} \right)^2 \)
\( (h + 3)^2 + k^2 = \frac{(h + 5)^2}{1} \)
\( h^2 + 6h + 9 + k^2 = h^2 + 10h + 25 \)
Now, bring all terms involving \(h\) to the right side and \(k^2\) to the left:
\( k^2 = h^2 + 10h + 25 - h^2 - 6h - 9 \)
\( k^2 = 4h + 16 \)
So, the locus of point \(P(h, k)\) is \( y^2 = 4x + 16 \). This is the required equation of the parabola.
In simple words: For a parabola, any point on its curve is equally far from a special point (the focus) and a special line (the directrix). We use distance formulas to write this rule as an equation.

🎯 Exam Tip: Remember to square both sides of the SP=PM equation to avoid square roots, and then carefully expand and collect terms to find the general equation of the parabola. Pay close attention to signs and coefficients.

Question 2. Find the vertex, axis, focus and latus rectum for following parabola:
(i) y² = 8x + 8y
(ii) x² + 2y = 8x - 7
Answer:
(i) Given equation: \( y^2 = 8x + 8y \)
Rearrange the terms to complete the square for \(y\):
\( y^2 - 8y = 8x \)
Add \( (-8/2)^2 = 16 \) to both sides:
\( y^2 - 8y + 16 = 8x + 16 \)
\( (y - 4)^2 = 8(x + 2) \)
This equation is in the standard form \( Y^2 = 4aX \), where \( Y = y - 4 \) and \( X = x + 2 \).
Comparing \( Y^2 = 8X \) with \( Y^2 = 4aX \), we find \( 4a = 8 \), so \( a = 2 \).
Now, let's find the required elements in the \( (X, Y) \) system and then convert them back to the \( (x, y) \) system.
Vertex:
In the \( (X, Y) \) system, the vertex is \( (0, 0) \).
So, \( X = 0 \implies x + 2 = 0 \implies x = -2 \)
And \( Y = 0 \implies y - 4 = 0 \implies y = 4 \)
Thus, the coordinates of the vertex are \( (-2, 4) \).
Axis:
In the \( (X, Y) \) system, the axis of the parabola \( Y^2 = 4aX \) is \( Y = 0 \).
So, \( y - 4 = 0 \implies y = 4 \)
Thus, the equation of the axis is \( y = 4 \).
Focus:
In the \( (X, Y) \) system, the focus is \( (a, 0) = (2, 0) \).
So, \( X = 2 \implies x + 2 = 2 \implies x = 0 \)
And \( Y = 0 \implies y - 4 = 0 \implies y = 4 \)
Thus, the coordinates of the focus are \( (0, 4) \).
Length of Latus Rectum:
The length of the latus rectum is \( 4a \).
Since \( a = 2 \), the length of the latus rectum is \( 4 \times 2 = 8 \).
(ii) Given equation: \( x^2 + 2y = 8x - 7 \)
Rearrange terms to complete the square for \(x\):
\( x^2 - 8x = -2y - 7 \)
Add \( (-8/2)^2 = 16 \) to both sides:
\( x^2 - 8x + 16 = -2y - 7 + 16 \)
\( (x - 4)^2 = -2y + 9 \)
This equation is in the standard form \( X^2 = 4aY \), where \( X = x - 4 \) and \( Y = y - \frac{9}{2} \).
Comparing \( X^2 = -2Y \) (for the transformed origin) with \( X^2 = 4aY \), we find \( 4a = -2 \), so \( a = -\frac{1}{2} \).
Let's find the required elements in the \( (X, Y) \) system and then convert them back to the \( (x, y) \) system.
Vertex:
In the \( (X, Y) \) system, the vertex is \( (0, 0) \).
So, \( X = 0 \implies x - 4 = 0 \implies x = 4 \)
And \( Y = 0 \implies y - \frac{9}{2} = 0 \implies y = \frac{9}{2} \)
Thus, the coordinates of the vertex are \( \left(4, \frac{9}{2}\right) \).
Axis:
In the \( (X, Y) \) system, the axis of the parabola \( X^2 = 4aY \) is \( X = 0 \).
So, \( x - 4 = 0 \implies x = 4 \)
Thus, the equation of the axis is \( x = 4 \).
Focus:
In the \( (X, Y) \) system, the focus is \( (0, a) = \left(0, -\frac{1}{2}\right) \).
So, \( X = 0 \implies x - 4 = 0 \implies x = 4 \)
And \( Y = -\frac{1}{2} \implies y - \frac{9}{2} = -\frac{1}{2} \implies y = \frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4 \)
Thus, the coordinates of the focus are \( (4, 4) \).
Length of Latus Rectum:
The length of the latus rectum is \( |4a| \).
Since \( a = -\frac{1}{2} \), the length of the latus rectum is \( |4 \times (-\frac{1}{2})| = |-2| = 2 \).
In simple words: To find the parts of a parabola, first rewrite its equation into a standard form like \( (y-k)^2 = 4a(x-h) \) or \( (x-h)^2 = 4a(y-k) \). Then, you can easily read off the vertex \((h,k)\), the value of \(a\), and use \(a\) to find the focus and latus rectum.

🎯 Exam Tip: Always complete the square for the squared variable (either \(x\) or \(y\)) to get the equation into standard form. Remember to shift the coordinates back to the original \( (x, y) \) system for the final answer.

Question 3. Length of double ordinate of parabola y² = 4ax is 8a. Prove that lines joining the origin to double ordinate will be perpendicular to each other.
Answer:
Let the equation of the parabola be \( y^2 = 4ax \).
A double ordinate is a chord perpendicular to the axis of the parabola. Let the coordinates of the endpoints of the double ordinate be \( P(x_1, y_1) \) and \( P'(x_1, -y_1) \).
Given that the length of the double ordinate is \( 8a \).
The length of the double ordinate is \( |y_1 - (-y_1)| = |2y_1| \).
So, \( 2y_1 = 8a \implies y_1 = 4a \).
Since point \( P(x_1, y_1) \) lies on the parabola \( y^2 = 4ax \), substitute \( y_1 = 4a \):
\( (4a)^2 = 4ax_1 \)
\( 16a^2 = 4ax_1 \)

\( \implies x_1 = \frac{16a^2}{4a} \)

\( \implies x_1 = 4a \)
So, the coordinates of \( P \) are \( (4a, 4a) \) and \( P' \) are \( (4a, -4a) \).
Now, we need to find the slopes of the lines joining the origin \( O(0, 0) \) to these points.
Slope of line \( OP \) (\(m_1\)) where \( O(0,0) \) and \( P(4a, 4a) \):
\( m_1 = \frac{4a - 0}{4a - 0} = \frac{4a}{4a} = 1 \)
Slope of line \( OP' \) (\(m_2\)) where \( O(0,0) \) and \( P'(4a, -4a) \):
\( m_2 = \frac{-4a - 0}{4a - 0} = \frac{-4a}{4a} = -1 \)
Now, let's check the product of the slopes:
\( m_1 \cdot m_2 = (1) \cdot (-1) = -1 \)
Since the product of their slopes is \( -1 \), the lines \( OP \) and \( OP' \) are perpendicular to each other. This proves the statement.
In simple words: When you connect the start point (origin) to the top and bottom ends of a special line segment (double ordinate) on a parabola, these two connecting lines will always meet at a right angle.

🎯 Exam Tip: To prove lines are perpendicular, always calculate their slopes and show that the product of the slopes is -1. Remember the formula for the length of a double ordinate and how to find the coordinates of its endpoints.

Question 4. If vertex and focus of parabola are at a distance of a and a' from origin to x-axis, then prove that equation of parabola will be y²= 4(a' – a)(x – a).
Answer:
Let the vertex of the parabola be \( V(a, 0) \) and the focus be \( F(a', 0) \), since both are on the x-axis at distances \( a \) and \( a' \) from the origin.
The distance between the vertex and the focus is \( |a' - a| \). Let this distance be \( d \).
In a parabola, the distance from the vertex to the focus is \( a_{parabola} \) (not to be confused with the given 'a' for vertex coordinate). So, \( d = |a' - a| \).
Since the focus \( (a', 0) \) is to the right of the vertex \( (a, 0) \), the parabola opens to the right. Therefore, \( a' > a \), and the parameter \( a_{parabola} = a' - a \).
The standard equation of a parabola with vertex \( (h, k) \) and opening to the right is \( (y - k)^2 = 4a_{parabola}(x - h) \).
Here, the vertex is \( (h, k) = (a, 0) \).
Substituting these values into the standard equation:
\( (y - 0)^2 = 4(a' - a)(x - a) \)
\( y^2 = 4(a' - a)(x - a) \)
This is the required equation of the parabola.
In simple words: If you know where a parabola's turning point (vertex) and special point (focus) are on the X-axis, you can use their distances from the start point (origin) to write down the parabola's full equation.

🎯 Exam Tip: When the vertex and focus are on the x-axis, the parabola opens either left or right. The value of 'a' in the standard form \(y^2 = 4a(x-h)\) is the distance between the vertex and the focus, \(a_{parabola}\), not the x-coordinate 'a' of the vertex.

Question 5. PQ is double ordinate of a parabola. Find the locus of its trisection point.
Answer:
Let the equation of the parabola be \( y^2 = 4ax \).
Let PQ be a double ordinate. Let it intersect the axis of the parabola (x-axis) at point A.
Let the coordinates of point \( A \) be \( (h_A, 0) \). Then the coordinates of \( P \) and \( Q \) are \( (h_A, y_P) \) and \( (h_A, -y_P) \) respectively.
Let \(R(h, k)\) be a trisection point of the double ordinate PQ. For simplicity, let R be the point closer to P than Q, so \( R \) divides \( PQ \) in the ratio \( 1:2 \). Or let \(R\) be a generic trisection point. The problem refers to "its trisection point", implying we find the locus of *any* such point. If \( R \) is a point that trisects \( PQ \), then it could be \( R \) or \( S \). Let's follow the provided solution's setup where \( R \) is above A and \( S \) is below A, and \( R(h,k) \) is the point whose locus we are finding.
Let \( A = (h_A, 0) \).
Since \( R(h, k) \) is a point on the ordinate, its x-coordinate is \( h_A \), so \( h = h_A \).
Let \( AR = k \). The points that trisect \( PQ \) mean that \( PR = RS = SQ \).
The double ordinate \( PQ \) is symmetric about the x-axis, with \( A \) being the midpoint of \( PQ \).
So, \( PA = AQ \).
If \( R \) is a trisection point such that \( PR = \frac{1}{3} PQ \), and \( A \) is the midpoint, then \( AR = PA - PR = PA - \frac{1}{3} PQ = PA - \frac{1}{3}(2PA) = PA - \frac{2}{3}PA = \frac{1}{3}PA \). This implies \( PA = 3AR \).
So, if \( AR = k \), then \( PA = 3k \).
Thus, the coordinates of \( P \) are \( (h, 3k) \).
Since \( P(h, 3k) \) lies on the parabola \( y^2 = 4ax \), we substitute its coordinates into the equation:
\( (3k)^2 = 4ah \)
\( 9k^2 = 4ah \)
Replacing \( h \) with \( x \) and \( k \) with \( y \) (to find the locus), we get:
\( 9y^2 = 4ax \)
This is the required locus of the trisection point.
In simple words: Imagine a straight line cutting across a parabola, passing through the x-axis. If you take a point on this line that is one-third of the way from either end (a trisection point), all such points will form another, narrower parabola.

🎯 Exam Tip: When finding the locus of a point, assign it general coordinates (h, k). Use the given conditions to form a relationship between h and k. Finally, replace h with x and k with y to get the equation of the locus.

Question 6. Prove that locus of mid points of all the chords passing through the vertex of parabola y2 = 4ax is a parabola.
Answer:
Let the equation of the parabola be \( y^2 = 4ax \). The vertex of this parabola is at the origin \( O(0,0) \).
Let \( P(at^2, 2at) \) be a general point on the parabola using parametric coordinates.
A chord passes through the vertex \( O(0,0) \) and point \( P(at^2, 2at) \).
Let \( M(h, k) \) be the midpoint of the chord \( OP \).
Using the midpoint formula for \( O(0,0) \) and \( P(at^2, 2at) \):
\( h = \frac{0 + at^2}{2} \implies h = \frac{at^2}{2} \)
\( k = \frac{0 + 2at}{2} \implies k = at \)
Now, we need to eliminate the parameter \( t \) to find the relationship between \( h \) and \( k \).
From the equation for \( k \), we have \( t = \frac{k}{a} \).
Substitute this value of \( t \) into the equation for \( h \):
\( h = \frac{a}{2} \left( \frac{k}{a} \right)^2 \)
\( h = \frac{a}{2} \left( \frac{k^2}{a^2} \right) \)
\( h = \frac{k^2}{2a} \)
Now, rearrange the equation to express \( k^2 \) in terms of \( h \):
\( k^2 = 2ah \)
Replacing \( h \) with \( x \) and \( k \) with \( y \) to represent the locus:
\( y^2 = 2ax \)
This is the equation of a parabola. This parabola has its vertex at the origin \( (0,0) \) and opens to the right, but it is "narrower" than the original parabola \( y^2=4ax \) because its coefficient of \(x\) is \(2a\) instead of \(4a\).
In simple words: If you take all the middle points of lines drawn from the parabola's tip (vertex) to any other point on the parabola, these middle points will also form a new parabola, which is half the "width" of the first one.

🎯 Exam Tip: When finding the locus of a midpoint of a chord through the origin, use parametric coordinates for points on the parabola. Apply the midpoint formula and then eliminate the parameter to get the equation of the locus.