NCERT Solutions Class 11 Mathematics Conic Sections Exercise 11.2

Question 1:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y² = 12x.

Answer:
The given equation is y² = 12x.
Here, the coefficient of x is positive. Hence, the parabola opens towards the right.
On comparing this equation with y² = 4ax, we obtain
    4a = 12  
=> a = 3
So, the coordinates of the focus = (a, 0) = (3, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of direcctrix, x = –a i.e., x = – 3 i.e., x + 3 = 0
Length of latus rectum = 4a = 4 * 3 = 12

Question 2:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x² = 6y.

Answer:
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 = 6/4
=> a = 3/2
So, the coordinates of the focus = (0, a) = (0, 3/2)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix, y = -a i.e. y = -3/2
Length of latus rectum = 4a = 4 * 3/2 = 2 * 3 = 6

Question 3:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y² = – 8x.
Answer :
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  
=> 4a = 8
=> a = 2
So, the 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, x = a i.e., x = 2
Length of latus rectum = 4a = 8

Question 4:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x² = – 16y.

Answer:
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
=> 4a = 16
=> a = 4
So, the 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, y = a i.e., y = 4
Length of latus rectum = 4a = 16

Question 5:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for y² = 10x.

Answer:
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 = 10/4
=> a = 5/2
Coordinates of the focus = (a, 0)
Since the given equation involves y², the axis of the parabola is the x-axis.
Equation of directrix, x = -a i.e. x = -5/2
Length of latus rectum = 4a = 10 

Question 6:
Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for x² = –9y.

Answer:
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
=> 4a = 9
=> a = -9/4
So, the coordinates of the focus = (0, -a) = (0, -9/4)
Since the given equation involves x², the axis of the parabola is the y-axis.
Equation of directrix, y = a i.e. y = 9/4
Length of latus rectum = 4a = 9

Question 7:
Find the equation of the parabola that satisfies the following conditions: Focus (6, 0);  directrix x = –6.

Answer:
Given, Focus is (6, 0) and the directrix, x = –6
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
Therefore, 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 the y-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     Thus, the equation of the parabola is y² = 24x.

Question 8:
Find the equation of the parabola that satisfies the following conditions: Focus (0, –3); directrix y = 3.

Answer:
Given, Focus = (0, –3) and the 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.