Inter 1st Year Maths Conic Sections Solutions Exercise 10b

Practicing the AP Board Solutions Class 11 Maths and Chapter 10 Inter 1st Year Maths Conic Sections Solutions Exercise 10b Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Conic Sections Solutions Exercise 10b

Conic Sections Exercise 10b Solutions

Conic Sections Class 11 Exercise 10b Solutions – Conic Sections 10b Exercise Solutions

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

Question 1.
y² = 12x
Solution:
Given y² = 12x
⇒ 4a = 12
⇒ a = 3
Vertex at (0, 0)
Focus = (a, 0) = (3, 0)
Axis y = 0
Directrix x = -a = -3
Length of latus rectum = 4a
= 4 × 3
= 12

Question 2.
x² = 6y
Solution:
The given equation is x² = 6y
Here, the coefficient of y is positive.
Hence, the parabola opens upwards.
On comparing this equation with x² = 4ay,
we obtain 4a = 6
⇒ a = \(\frac {3}{2}\)
∴ Co-ordinates of the focus = (0, a) = (0, \(\frac {3}{2}\))
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix is y = -a
i.e.,y = \(-\frac {3}{2}\)
∴ Length of latus rectum = 4a = 6.

Question 3.
y² = -8x
Solution:
The given equation is y² = -8x
Here, the coefficient of x is negative.
Hence, the parabola opens towards the left.
On comparing this equation with y² = -4ax,
we obtain -4a = – 8
⇒ a = 2
∴ Coordinates of the focus = (-a, 0) = (-2, 0)
Since the given equation involves y²,
The axis of the parabola is the x-axis.
Equation of directrix is x = a
i.e., x = 2
∴ Length of latus rectum = 4a = 8.

Question 4.
x² = -16y
Solution:
The given equation is x² = -16y
Here, the coefficient of y is negative.
Hence, the parabola opens downwards.
On comparing this equation with x² = -4ay,
We obtain -4a = -16
⇒ a = 4
∴ Coordinates of the focus = (0, -a) = (0, -4)
Since the given equation involves x²,
The axis of the parabola is the y-axis.
Equation of directrix is y = a
i.e., y = 4
∴ Length of latus rectum = 4a = 16.

Question 5.
y² = 10x
Solution:
The given equation is y² = 10x
Here, the coefficient of x is positive.
Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax,
we obtain 4a = 10
⇒ a = \(\frac {5}{2}\)
∴ Coordinates of the focus = (a, 0) = (\(\frac {5}{2}\), 0)
Length of latus rectum = 4a = 10.
Since the given equation involves y²
The axis of the parabola is the x-axis.
Equation of directrix is x = -a
i.e., x = \(-\frac {5}{2}\)

Question 6.
x² = -9x
Solution:
The given equation is x² = -9y
Here, the coefficient of y is negative.
Hence, the parabola opens downwards.
On comparing this equation with x² = -4ay,
We obtain -4a = -9
⇒ a = \(\frac {9}{4}\)
∴ Co-ordinates of the focus = (0, -a) = (0, \(-\frac {9}{4}\))
Since the given equation involves x²
The axis of the parabola is the y-axis.
Equation of directrix is y = a
i.e., y = \(\frac {9}{4}\)
Length of latus rectum = 4a = 9.

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
Solution:
Focus (6, 0); Directrix x = 6
Midpoint between x = 6 and x = -6
x = \(\frac{6+(-6)}{2}\) = 0, y = 0
⇒ Vertex at (0, 0)
Since the focus is at (6, 0), the Vertex is at (0, 0)
So a = 6
⇒ y² = 4(6)x
⇒ y² = 24x

Question 8.
Focus (0, 3); directrix y = 3
Solution:
Focus = (0, -3); directrix y = 3.
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
Therefore, the equation of the parabola is either of the form x² = 4ay or x² = -4ay.
It is also seen that the directrix, y = 3, is above the x-axis, while the focus (0, -3) is below the x-axis.
Hence, the parabola is of the form x² = -4ay.
Here a = 3
Thus, the equation of the parabola is x² = -12y.

Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Vertex (0, 0); focus (3, 0)
Since the vertex of the parabola is (0, 0) and the focus lies on the positive x-axis, the x-axis is the axis of the parabola.
while the equation of the parabola is of the form y² = 4ax.
Since the focus is (3, 0), a = 3.
Thus, the equation of the parabola is y² = 4 × 3 × x
⇒ y² = 12x

Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Since the vertex is at (0, 0) and the focus is at (-2, 0,) which lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, the equation of the parabola is of the form y² = 4ax
Thus, the equation of the parabola is y² = 4(-2)x
⇒ y² = -8x

Question 11.
Vertex (0, 0) passing through (2, 3), and the axis is along the x-axis.
Solution:
Since the vertex is (0, 0) and the axis of the parabola is the x-axis,
The equation of the parabola is either of the form y² = 4ax or y² = -4ax.
The parabola passes through the point (2, 3), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form y² = 4ax,
while point (2, 3) must satisfy the equation y² = 4ax.
∴ 3² = 4a(2)
⇒ a = \(\frac {9}{8}\)
Thus, the equation of the parabola is
y² = 4(\(\frac {9}{8}\))x
⇒ y² = \(\frac {9}{2}\)x
⇒ 2y² = 9x

Question 12.
Vertex (0, 0) passing through (5, 2) and symmetric with respect to the y-axis.
Solution:
Since the vertex is (0, 0) and the parabola is symmetric about the y-axis,
the equation of the parabola is either of the form x² = 4ay or x² – 4ay.
The parabola passes through the point (5, 2), which lies in the first quadrant.
Therefore, the equation of the parabola is of the form x² = 4ay,
while point (5, 2) must satisfy the equation x² = 4ay.
∴ (5)² = 4 × a × 2
⇒ 25 = 8a
⇒ a = \(\frac {25}{8}\)
Thus, the equation of the parabola is x2 = 4(\(\frac {25}{8}\))y
⇒ 2x² = 25y

II.

Question 1.
Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).
Solution:
S = (3, 5), A = (1, 3)
Let Z(α, β) be the point of intersection of the axis and the directrix.
∴ A is the mid point of \(\overline{\mathrm{SZ}}\)
⇒ (1, 3) = \(\left(\frac{3+\alpha}{2}, \frac{5+\beta}{2}\right)\)
⇒ α = -1, β = 1
∴ Z (-1, 1)
Slope of \(\overline{\mathrm{SZ}}\) is \(\frac{1-5}{-1-3}\) = 1
⇒ Slope of the directrix is -1.
Equation of the directrix is x + y = 0
Let P(x, y) be a point on the parabola.
P is a point on the parabola
⇒ SP = PM
⇒ SP² = PM²
⇒ (x – 3)² + (y – 5)² = \(\frac{(x+y)^2}{1^2+1^2}\)
⇒ 2(x² – 6x + 9 + y² – 10y + 25) = x² + 2xy + y²
⇒ x² – 2xy + y² – 12x – 20y + 68 = 0
∴ Equation of the parabola is x² – 2xy + y² – 12x – 20y + 68 = 0.

Question 2.
Find the equation of the parabola whose focus is (-2, 3), and whose directrix is the line 2x + 3y – 4 = 0. Also, find the length of the latus rectum and the equation of the axis of the parabola.
Solution:
Focus S = (-2, 3)
Directrix is 2x + 3y – 4 = 0.
Let P(x, y) be a point on the parabola
P is a point on the parabola
⇒ SP = PM
⇒ SP² = PM²
⇒ (x + 2)² + (y – 3)² = \(\frac{(2 x+3 y-4)^2}{2^2+3^2}\)
⇒ 13(x² + 4x + 4 + y² – 6y + 9) = 4x² + 12xy + 9y² – 16x – 24y + 16
⇒ 9x² – 12xy + 4y² + 68x – 54y + 153 = 0
∴ Equation of the parabola is 9x² – 12xy + 4y² + 68x – 54y + 153 = 0
Length of latus rectum = 4a
= 2(2a)
= 2(the perpendicular distance from focus to directrix)
= \(\frac{2|2(-2)+3(3)-4|}{\sqrt{4+9}}\)
= \(\frac{2}{\sqrt{13}}\)
Slope of the directrix is \(\frac {-2}{3}\)
⇒ Slope of the axis is \(\frac {3}{2}\)
Equation of the axis is y – 3 = \(\frac {3}{2}\)(x + 2)
⇒ 2y – 6 = 3x + 6
⇒ 3x – 2y + 12 = 0

Question 3.
Find the equation of the parabola whose axis is parallel to the x-axis and which passes through the points (-2, 1), (1, 2), and (-1, 3).
Solution:
Let the equation of the parabola be x = ly² + my + n
Since (-2, 1) lies on the parabola,
-2 = l + m + n
⇒ l + m + n = -2 ……(1)
Since (1, 2) lies on the parabola,
1 = 4l + 2m + n
⇒ 4l + 2m + n = 1 ……(2)
Since (-1, 3) lies on the parabola,
-1 = 9l + 3m + n
⇒ 9l + 3m + n = -1 ………..(3)
(2) – (1) ⇒ 3l + m = 3 ……….(4)
(3) – (2) ⇒ 5l + m = -2 ……(5)
(3) – (4) ⇒ 2l = -5 ⇒ l = \(\frac {-5}{2}\)
(4) ⇒ \(\frac {-15}{2}\) + m = 3
⇒ m = \(\frac {21}{2}\)
(1) ⇒ \(-\frac{5}{2}+\frac{21}{2}\) + n = -2
∴ Equation of the parabola is x = \(-\frac{5}{2} y^2+\frac{21}{2} y-10\)
⇒ 2x = -5y² + 21y – 20
⇒ 5y² + 2x – 21y + 20 = 0

Question 4.
Find the equation of the parabola whose axis is parallel to the y-axis and which passes through the points (4, 5), (-2, 11), and (-4, 21).
Solution:
Let the equation of the parabola whose axis is parallel to the y-axis be
x² + ax + by + c = 0 ……….(1)
Since it passes through (4, 5), (-2, 11), (-4, 21), we have
4a + 5b + c = -16 ……..(2)
-2a + 11b + c = -4 ……….(3)
-4a + 42b + c = -16 ………(4)
Solving (2), (3), (4)
we get a = -4, b = -2, c = 10
Equation of the parabola is x² – 4x – 2y + 10 = 0.