GSEB Class 11 Maths Solutions Chapter 11 Conic Sections Exercise 11.2

In each of the following questions 1 to 6, find the co-ordinates of the focus, axis of the parabola, the equation of the directrix and length of latus rectum:

Question 1. \( y^{2} = 12x \)
Answer: The given equation of parabola is \( y^{2} = 12x \).
\( \implies \) When we compare this with \( y^{2} = 4ax \), we get \( 4a = 12 \).
\( \implies a = 3 \).
The focus coordinates are \( (a, 0) \), which gives \( (3, 0) \).
The axis is the x-axis, so its equation is \( y = 0 \).
The directrix equation is \( x = -a \), which means \( x = -3 \).
The length of the latus rectum is \( 4a = 4(3) = 12 \).
In simple words: For the parabola \( y^{2} = 12x \), the focus is at (3,0), the x-axis is its main line, and the directrix is the line \( x = -3 \). Its latus rectum has a length of 12.

Exam Tip: Remember that for parabolas of the form \( y^2 = 4ax \), the focus is \( (a,0) \), axis is \( y=0 \), and directrix is \( x=-a \). Always clearly state the values of 'a' first.

Question 2. \( x^{2} = 6y \)
Answer: The given equation of parabola is \( x^{2} = 6y \).
\( \implies \) When we compare this with \( x^{2} = 4ay \), we find \( 4a = 6 \).
\( \implies a = \frac{6}{4} = \frac{3}{2} \).
The focus coordinates are \( (0, a) \), which means \( (0, \frac{3}{2}) \).
The axis is the y-axis, so its equation is \( x = 0 \).
The directrix equation is \( y = -a \), which gives \( y = - \frac{3}{2} \).
The length of the latus rectum is \( 4a = 6 \).
In simple words: For \( x^{2} = 6y \), the focus is at \( (0, \frac{3}{2}) \), the y-axis is its main line, and the directrix is the line \( y = - \frac{3}{2} \). Its latus rectum has a length of 6.

Exam Tip: When the parabola equation is \( x^2 = 4ay \), it opens upwards. The focus is on the positive y-axis, and the directrix is a horizontal line below the x-axis.

Question 3. \( y^{2} = -8x \)
Answer: The given equation of parabola is \( y^{2} = -8x \).
\( \implies \) When we compare this with \( y^{2} = -4ax \), we get \( 4a = 8 \).
\( \implies a = 2 \).
The focus coordinates are \( (-a, 0) \), which means \( (-2, 0) \).
The axis is the x-axis, so its equation is \( y = 0 \).
The directrix equation is \( x = a \), which gives \( x = 2 \).
The length of the latus rectum is \( 4a = 8 \).
In simple words: For the parabola \( y^{2} = -8x \), the focus is at (-2,0), the x-axis is its main line, and the directrix is the line \( x = 2 \). Its latus rectum has a length of 8.

Exam Tip: Pay close attention to the sign in the parabola equation. \( y^2 = -4ax \) opens to the left, which means its focus is on the negative x-axis and the directrix is on the positive x-axis.

Question 4. \( x^{2} = -16y \)
Answer: The given equation of parabola is \( x^{2} = -16y \).
\( \implies \) When we compare this with \( x^{2} = -4ay \), we find \( 4a = 16 \).
\( \implies a = 4 \).
The focus coordinates are \( (0, -a) \), which means \( (0, -4) \).
The axis is the y-axis, so its equation is \( x = 0 \).
The directrix equation is \( y = a \), which gives \( y = 4 \).
The length of the latus rectum is \( 4a = 16 \).
In simple words: For \( x^{2} = -16y \), the focus is at \( (0, -4) \), the y-axis is its main line, and the directrix is the line \( y = 4 \). Its latus rectum has a length of 16.

Exam Tip: When the parabola equation is \( x^2 = -4ay \), it opens downwards. The focus is on the negative y-axis, and the directrix is a horizontal line above the x-axis.

Question 5. \( y^{2} = 10x \)
Answer: The given equation of parabola is \( y^{2} = 10x \).
\( \implies \) When we compare this with \( y^{2} = 4ax \), we find \( 4a = 10 \).
\( \implies a = \frac{10}{4} = \frac{5}{2} \).
The focus coordinates are \( (a, 0) \), which means \( (\frac{5}{2}, 0) \).
The axis is the x-axis, so its equation is \( y = 0 \).
The directrix equation is \( x = -a \), which gives \( x = - \frac{5}{2} \).
The length of the latus rectum is \( 4a = 10 \).
In simple words: For \( y^{2} = 10x \), the focus is at \( (\frac{5}{2}, 0) \), the x-axis is its main line, and the directrix is the line \( x = - \frac{5}{2} \). Its latus rectum has a length of 10.

Exam Tip: Make sure to correctly identify the value of 'a' by setting \( 4a \) equal to the coefficient of the linear term in the parabola equation.

Question 6. \( x^{2} = -9y \)
Answer: The given equation of parabola is \( x^{2} = -9y \).
\( \implies \) When we compare this with \( x^{2} = -4ay \), we get \( 4a = 9 \).
\( \implies a = \frac{9}{4} \).
The focus coordinates are \( (0, -a) \), which means \( (0, - \frac{9}{4}) \).
The axis is the y-axis, so its equation is \( x = 0 \).
The directrix equation is \( y = a \), which gives \( y = \frac{9}{4} \).
The length of the latus rectum is \( 4a = 9 \).
In simple words: For \( x^{2} = -9y \), the focus is at \( (0, - \frac{9}{4}) \), the y-axis is its main line, and the directrix is the line \( y = \frac{9}{4} \). Its latus rectum has a length of 9.

Exam Tip: Remember that when the \( x^2 \) term is present, the parabola opens either up or down. A negative coefficient on the \( y \) term indicates it opens downwards.

In each of the following questions 7 to 12, find equation of the parabola that satisfies the given conditions:

Question 7. Focus is (6, 0) and directrix is \( x = -6 \).
Answer: The focus is \( F(6, 0) \) and the directrix is \( x = -6 \).
\( \implies \) The vertex is the midpoint of the focus and the point where the directrix intersects the axis. The point on the directrix (which is a vertical line \( x = -6 \)) that is closest to the focus (6,0) is (-6,0).
\( \implies \) So, the vertex is the midpoint of \( (6, 0) \) and \( (-6, 0) \).
Vertex \( = (\frac{6 + (-6)}{2}, \frac{0+0}{2}) = (0, 0) \).
Since the focus is \( (6, 0) \), we have \( a = 6 \).
Because the focus is on the positive x-axis and the directrix is \( x = -a \), the equation of the parabola is of the form \( y^{2} = 4ax \).
\( \implies \) Substituting \( a = 6 \) into the equation, we get \( y^{2} = 4(6)x \).
\( \implies y^{2} = 24x \).
In simple words: With focus (6,0) and directrix \( x = -6 \), the vertex is (0,0). Since \( a = 6 \) and it opens right, the parabola's equation is \( y^{2} = 24x \).

Exam Tip: If the focus is \( (a, 0) \) and the directrix is \( x = -a \), the vertex is always at the origin \( (0,0) \), and the parabola opens to the right with equation \( y^2 = 4ax \).

Question 8. Focus is \( (0, -3) \), directrix \( y = 3 \).
Answer: The focus is \( F(0, -3) \) and the directrix is \( y = 3 \).
\( \implies \) The vertex is the midpoint of the focus and the point where the directrix intersects the axis. The point on the directrix (which is a horizontal line \( y = 3 \)) that is closest to the focus (0,-3) is (0,3).
\( \implies \) So, the vertex is the midpoint of \( (0, -3) \) and \( (0, 3) \).
Vertex \( = (\frac{0+0}{2}, \frac{-3+3}{2}) = (0, 0) \).
Since the focus is \( (0, -3) \), we have \( a = 3 \).
Because the focus is on the negative y-axis and the directrix is \( y = a \), the equation of the parabola is of the form \( x^{2} = -4ay \).
\( \implies \) Substituting \( a = 3 \) into the equation, we get \( x^{2} = -4(3)y \).
\( \implies x^{2} = -12y \).
In simple words: The focus is \( (0, -3) \) and the directrix is \( y = 3 \). This means the vertex is \( (0, 0) \) and the value of 'a' is 3. Since the parabola opens downwards, its equation is \( x^{2} = -12y \).

Exam Tip: If the focus is \( (0, -a) \) and the directrix is \( y = a \), the vertex is at the origin \( (0,0) \), and the parabola opens downwards with equation \( x^2 = -4ay \).

Question 9. Vertex (0, 0) and Focus (3, 0).
Answer: The vertex is \( (0, 0) \) and the focus is \( (3, 0) \).
\( \implies \) Since the vertex is at the origin and the focus is at \( (3, 0) \), the parabola opens to the right along the x-axis.
\( \implies \) This form corresponds to \( y^{2} = 4ax \).
From the focus \( (3, 0) \), we get \( a = 3 \).
\( \implies \) Substituting \( a = 3 \) into the equation, we find \( y^{2} = 4(3)x \).
\( \implies y^{2} = 12x \).
In simple words: With its vertex at (0,0) and focus at (3,0), the parabola opens to the right. Therefore, its equation is \( y^{2} = 12x \).

Exam Tip: When the vertex is at the origin and the focus is on an axis, the equation form is determined by which axis the focus lies on and its sign.

Question 10. Vertex (0, 0) and Focus (- 2, 0).
Answer: The vertex is \( (0, 0) \) and the focus is \( (-2, 0) \).
\( \implies \) Since the vertex is at the origin and the focus is at \( (-2, 0) \), the parabola opens to the left along the negative x-axis.
\( \implies \) This form corresponds to \( y^{2} = -4ax \).
From the focus \( (-2, 0) \), we get \( a = 2 \).
\( \implies \) Substituting \( a = 2 \) into the equation, we find \( y^{2} = -4(2)x \).
\( \implies y^{2} = -8x \).
In simple words: With its vertex at (0,0) and focus at (-2,0), the parabola opens to the left. Thus, its equation is \( y^{2} = -8x \).

Exam Tip: A focus with a negative x-coordinate and a vertex at the origin indicates a parabola that opens left, matching the \( y^2 = -4ax \) form.

Question 11. Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Answer: The vertex is \( (0, 0) \) and the axis is along the x-axis.
\( \implies \) This means the parabola is of the form \( y^{2} = 4ax \) or \( y^{2} = -4ax \).
The parabola passes through the point \( (2, 3) \). Since \( y^{2} \) must be positive, we consider the form \( y^{2} = 4ax \).
Substituting \( x = 2 \) and \( y = 3 \) into the equation:
\( (3)^{2} = 4a(2) \)
\( 9 = 8a \)
\( \implies a = \frac{9}{8} \).
Now, substitute the value of \( a \) back into the equation \( y^{2} = 4ax \):
\( y^{2} = 4(\frac{9}{8})x \)
\( y^{2} = \frac{9}{2}x \)
\( \implies 2y^{2} = 9x \).
In simple words: The parabola starts at (0,0) and opens along the x-axis. Since it passes through (2,3), we put these values into \( y^{2} = 4ax \) to find \( a = \frac{9}{8} \). The final equation is \( 2y^{2} = 9x \).

Exam Tip: If the axis is the x-axis, the equation will be \( y^2 = 4ax \) (or \( -4ax \)). If the axis is the y-axis, the equation will be \( x^2 = 4ay \) (or \( -4ay \)). Use the given point to find 'a'.

Question 12. Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Answer: The vertex is \( (0, 0) \) and the parabola is symmetric with respect to the y-axis.
\( \implies \) This means the parabola is of the form \( x^{2} = 4ay \) or \( x^{2} = -4ay \).
The parabola passes through the point \( (5, 2) \). Since \( x^{2} \) must be positive, we consider the form \( x^{2} = 4ay \).
Substituting \( x = 5 \) and \( y = 2 \) into the equation:
\( (5)^{2} = 4a(2) \)
\( 25 = 8a \)
\( \implies a = \frac{25}{8} \).
Now, substitute the value of \( a \) back into the equation \( x^{2} = 4ay \):
\( x^{2} = 4(\frac{25}{8})y \)
\( x^{2} = \frac{25}{2}y \)
\( \implies 2x^{2} = 25y \).
In simple words: The parabola starts at (0,0) and opens along the y-axis because of symmetry. Since it passes through (5,2), we put these values into \( x^{2} = 4ay \) to find \( a = \frac{25}{8} \). The final equation is \( 2x^{2} = 25y \).

Exam Tip: Symmetry about the y-axis means the equation will have an \( x^2 \) term, while symmetry about the x-axis means it will have a \( y^2 \) term. Use the coordinates of the given point to find the parameter 'a'.