Chapter 11 Conic Sections Ex 11.2 Solutions
Question - 1 : - 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.
Answer - 1 : -
Given:
The equation is y2 =12x
Here we know that the coefficient of x is positive.
So, the parabola opens towards the right.
On comparing this equation with y2 =4ax, we get,
4a = 12
a = 3
Thus, the co-ordinates of the focus = (a, 0) = (3, 0)
Since, the given equation involves y2,the axis of the parabola is the x-axis.
∴ Theequation of directrix, x = -a, then,
x + 3 = 0
Length of latus rectum = 4a = 4 × 3 = 12
Question - 2 : - 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.
Answer - 2 : -
Given:
The equation is x2 =6y
Here we know that the coefficient of y is positive.
So, the parabola opens upwards.
On comparing this equation with x2 =4ay, we get,
4a = 6
a = 6/4
= 3/2
Thus, the co-ordinates of the focus = (0,a) = (0, 3/2)
Since, the given equation involves x2,the axis of the parabola is the y-axis.
∴ Theequation of directrix, y =-a, then,
y = -3/2
Length of latus rectum = 4a = 4(3/2) = 6
Question - 3 : - 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.
Answer - 3 : -
Given:
The equation is y2 =-8x
Here we know that the coefficient of x is negative.
So, the parabola open towards the left.
On comparing this equation with y2 =-4ax, we get,
-4a = -8
a = -8/-4 = 2
Thus, co-ordinates 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, x =a, then,
x = 2
Length of latus rectum = 4a = 4 (2) = 8
Question - 4 : - 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.
Answer - 4 : -
Given:
The equation is x2 =-16y
Here we know that the coefficient of y is negative.
So, the parabola opens downwards.
On comparing this equation with x2 =-4ay, we get,
-4a = -16
a = -16/-4
= 4
Thus, co-ordinates of the focus = (0,-a) = (0,-4)
Since, the given equation involves x2,the axis of the parabola is the y-axis.
∴ Theequation of directrix, y =a, then,
y = 4
Length of latus rectum = 4a = 4(4) = 16
Question - 5 : - 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.
Answer - 5 : -
Given:
The equation is y2 =10x
Here we know that the coefficient of x is positive.
So, the parabola open towards the right.
On comparing this equation with y2 =4ax, we get,
4a = 10
a = 10/4 = 5/2
Thus, co-ordinates of the focus = (a,0) = (5/2, 0)
Since, the given equation involves y2,the axis of the parabola is the x-axis.
∴ Theequation of directrix, x =-a, then,
x = – 5/2
Length of latus rectum = 4a = 4(5/2) = 10
Question - 6 : - 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.
Answer - 6 : -
Given:
The equation is x2 =-9y
Here we know that the coefficient of y is negative.
So, the parabola open downwards.
On comparing this equation with x2 =-4ay, we get,
-4a = -9
a = -9/-4 = 9/4
Thus, co-ordinates of the focus = (0,-a) = (0, -9/4)
Since, the given equation involves x2,the axis of the parabola is the y-axis.
∴ Theequation of directrix, y = a, then,
y = 9/4
Length of latus rectum = 4a = 4(9/4) = 9
Question - 7 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Focus (6,0); directrix x = – 6
Answer - 7 : -
Focus (6,0) and directrix x = -6
We know that the focus lies on the x–axis is the axis of theparabola.
So, the equation of the parabola is either of the form y2 = 4ax or y2 =-4ax.
It is also seen that the directrix, x = -6 is to the left of they- axis,
While the focus (6, 0) is to the right of the y –axis.
Hence, the parabola is of the form y2 =4ax.
Here, a = 6
∴ Theequation of the parabola is y2 =24x.
Question - 8 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Focus (0,–3); directrix y = 3
Answer - 8 : -
Given:
Focus (0, -3) and directrix y = 3
We know that the focus lies on the y–axis, the y-axis is theaxis of the parabola.
So, 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
∴ Theequation of the parabola is x2 =-12y.
Question - 9 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Vertex (0, 0); focus (3, 0)
Answer - 9 : -
Given:
Vertex (0, 0) and focus (3, 0)
We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]
The equation of the parabola is of the form y2 = 4ax.
Since, the focus is (3, 0), a = 3
∴ Theequation of the parabola is y2 =4 × 3 × x,
y2 = 12x
Question - 10 : - In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:Vertex (0, 0); focus (–2, 0)
Answer - 10 : -
Given:
Vertex (0, 0) and focus (-2, 0)
We know that the vertex of the parabola is (0, 0) and the focuslies on the positive x-axis. [x-axis is the axis of the parabola.]
The equation of the parabola is of the form y2=-4ax.
Since, the focus is (-2, 0), a = 2
∴ Theequation of the parabola is y2 =-4 × 2 × x,
y2 = -8x