RBSE Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2

Rajasthan Board RBSE Solutions for Class 11 Maths Chapter 11 Conic Sections Ex 11.2 Textbook Exercise Questions and Answers.

RBSE Class 11 Maths Solutions Chapter 11 Conic Sections Ex 11.2

In each of the following exercise 1 to 6, find the coordinates of the focus, axis of the prabola, the equation of the directrix and the length of the latus rectum.

Question 1.
y² = 12x
Answer:
Equation of the given parabola = y² = 12x
Comparing it with y² = 4ax
Latus rectum 4a = 12
Here, a = 3
Thus, coordinates of the focus of the parabola (a, 0) = (3, 0)
Since, in the equation of the parabola, one term is y², so, parabola will be symmetric about x-axis.
Thus, axis of parabola will be x-axis.
Equation of directrix x = - 3 (by x = - a)
Length of latus rectum = 4a = 4 × 3 = 12

Question 2.
x² = 6y
Answer:
Equation of the given parabola
x² = 6y
Comparing given equation of parabola x² = 6y by x² = 4ay
4a = 6
⇒ a = \(\frac{6}{4}\)
⇒ a = \(\frac{3}{2}\)
Thus, focus of parabola (0, a) = (0, \(\frac{3}{2}\))
Again, axis of parabola is y-axis.
Thus, equation of directrix y = - a
y = \(\frac{-3}{2}\)
⇒ 2y + 3 = 0
Length of latus rectum = 4a
= 4 × \(\frac{3}{2}\) = 6

Question 3.
y² = -8x
Answer:
Equation of given parabola
y² = - 8x
The given equation involves y²,
So Axis of parabola = X-axis
Comparing given equation of parabola by y² = 4ax
4a = -8
⇒ a = -2
Focus of parabola (a, 0) = (-2,0)
Equation of directrix x = - a
⇒ x = -(-2)
⇒ x - 2 = 0
Length of latus rectum = 4a
= 4 × (-2)
= -8 =8 units
[Length is never negative]

Question 4.
x²= - 16y
Answer:
Equation of given parabola
x² = -16 y
The given equation involves x²,
So Axis of parabola = Y-axis
Comparing, x² = -16y by x² = 4ay
4a = -16
⇒ a = -4
Focus of parabola = (0, 0) = (0, -4)
Equation of directrix = y + a = 0
⇒ y + (-4) = 0
⇒ y = 4
Length of latus rectum = 4a = 4 × 4 = 16units

Question 5.
y² = 10x
Answer:
Equation of given parabola y² = 10x
The equation involves y²
Axis of parabola = X-axis
Again, comparing y² = 10x by y² = 4 ax
4a = 10
⇒ a = \(\frac{10}{4}=\frac{5}{2}\)
Focus of parabola (a, 0) = (\(\frac{5}{2}\), 0)
Equation of directrix x + a = 0
⇒ x + \(\frac{5}{2}\) = 0
⇒ 2x + 5 = 0
Length of latus rectum = 4a = 4 × \(\frac{5}{2}\) = 10 units

Question 6.
x² = -9y
Answer:
Equation of given parabola x² = -9 y
The equation involves x²,
so Axis of parabola is y-axis (x = 0)
Again, comparing x² = -9y with x² = 4ay
4a = -9
a = \(\frac{-9}{4}\)
Focus of parabola = (0, a) = (0, \(\frac{-9}{4}\))
Equation of directrix y + a = 0
⇒ y - \(\frac{9}{4}\) = 0
⇒ 4y - 9 = 0
Length of latus rectum = 4a = 4 × \(\frac{9}{4}\) = 9 units

In each of the exercises 7 to 12, find the equation of the parabola that satisfies the given conditions.

Question 7.
Focus (6,0), directrix x = - 6.
Answer:
Let any point on parabola be p(x, y) then distance of focus (6, 0) from point (x, y).
= \(\sqrt{(x-6)^2+(y-0)^2}\)
and distance of directrix x = -6 from point (x, y)
= \(\frac{x+6}{\sqrt{1}}\) = x + 6
According to the definition of parabola,
Distance of focus (6,0) from point (x, y)
= Distance of directrix from point (x, y)
So, \(\sqrt{(x-6)^2+(y-0)^2}\) = x + 6
Squaring both sides, we have
(x - 6)² + y² = (x + 6)²
⇒ x² - 12x + 36 + y² = x² + 12x + 36
⇒ y² = x² + 12x + 36 + 12x - 36 - x²
⇒ y² = 24x
Thus, equation of parabola:
y² = 24x

Question 8.
Focus (0, -3), directrix y = 3.
Answer:
Let (x, y) be any point on parabola
Then distance of focus (0, -3) from point (x, y)
= Distance of directrix y = 3 from point (x, y)
\(\sqrt{(x-0)^2+(y+3)^2}=\frac{y-3}{\sqrt{1}}\) = y - 3
Squaring of both sides, we get
(x - 0)² + (y + 3)² = (y - 3)²
⇒ x² + y² + 6y + 9 = y² - 6y + 9
⇒ x² = y² - 6y + 9 - 6y - 9 - y²
⇒ x² = -12y
Thus equation of parabola : x² = -12y

Question 9.
Vertex, (0,0), focus = (3, 0)
Answer:
Distance between vertex (0, 0) and focus (3, 0)
= \(\sqrt{(3-0)^2+(0-0)^2}\) = 3
So, distance of directrix from vertex = 3 then equation of directrix x + 3 = 0
⇒ x = -3
Let (x, y) be any point on parabola.
By definition of parabola.
Distance of focus (3,0) from point (x, y)
= Distance of directrix x + 3 = 0 from point (x, y)
∴ \(\sqrt{(x-3)^2+(y+0)^2}=\frac{x+3}{\sqrt{1}}\) = x + 3
Squaring both sides, we get
(x - 3) + (y - 0)² = (x + 3)²
⇒ x² - 6x + 9 + y² = x² + 6x + 9
⇒ y² = x² + 6x + 9 - x² + 6x - 9
⇒ y² = 12x
Thus equation of parabola : y² = 12x

Question 10.
Vertex.(0, 0) focus = (- 2, 0)
Answer:
Distance between vertex (0,0) and focus (-2,0)
= \(\sqrt{(-2-0)^2+(0-0)^2}\) = 2
Distance of directrix from vertex = 2
Equation of directrix = x + a = 0
⇒ x - 2 = 0
⇒ x = 2
Let point (x, y) be any point on parabola, then distance of focus (-2, 0) from point (x, y)
= Distance of directrix x - 2 = 0 from point (x, y)
\(\sqrt{(x+2)^2+(y-0)^2}=\frac{x-2}{\sqrt{1}}\) = (x - 2)
Squaring both sides, we have
(x + 2)² + (y - 0)² = (x - 2)²
x² + 4x + 4 + y² = x² - 4x + 4
y² = x² - 4x + 4 - x² - 4x - 4
⇒ y² = -8x
Thus, equation of parabola is y² = -8x

Question 11.
Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
Answer:
We have,
Axis of parabola is along X-axis.
since, parabola y² = 4ax passes through point (2, 3)
So, 3² = 4a(2)
⇒ 9 = 8a
⇒ a = \(\frac{9}{8}\)
⇒ y = 4.\(\frac{9}{8}\)x
⇒ 2y² = 9x
Thus, equation of parabola is 2y² = 9x

Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect toy-axis.
Answer:
We have,
Parabola is symmetric about Y-axis.
Let equation of parabola is x² = 4ay
Parabola x² = 4ay passes through point (5, 2)
So, 5² = 4a × 2
⇒ 25 = 8a
⇒ a = \(\frac{25}{8}\)
Equation of parabola, x² = 4 × \(\frac{25}{8}\)y
⇒ x² = \(\frac{25}{2}\)y
⇒ 2x² = 25y
Thus, equation of parabola 2x² = 25y