NCERT Solutions for Class 11 Maths Chapter 10 Conic Sections Ex 10.2
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.
Ex 10.2 Class 11 Maths Question 1.
y2 = 12x
Solution:
The given equation of parabola is:
y2 = 12x, which is of the form y2 = 4ax.
∴ 4a = 12 ⇒ a = 3
∴ Coordinates of the focus are (3, 0).
Axis of parabola is y = 0.
Equation of the directrix is x = -3 ⇒ x + 3 = 0
Length of the latus rectum = 4 × 3 = 12.
Ex 10.2 Class 11 Maths Question 2.
x2 = 6y
Solution:
The given equation of parabola is:
x2 = 6y which is of the form x2 = 4ay.
Ex 10.2 Class 11 Maths Question 3.
y2 = –8x
Solution:
The given equation of parabola is:
y2 = -8x, which is of the form y2 = –4ax.
∴ 4a = 8 ⇒ a = 2
∴ Coordinates of the focus are (-2, 0).
Axis of parabola is y = 0.
Equation of the directrix is x = 2 ⇒ x – 2 = 0
Length of the latus rectum = 4 × 2 = 8.
Ex 10.2 Class 11 Maths Question 4.
x2 = -16y
Solution:
The given equation of parabola is:
x2 = -16y, which is of the form x2 = -4ay.
∴ 4a = 16 ⇒ a = 4
∴ Coordinates of the focus are (0, -4).
Axis of parabola is x = 0.
Equation of the directrix is y = 4 ⇒ y – 4 = 0
Length of the latus rectum = 4 × 4 = 16.
Ex 10.2 Class 11 Maths Question 5.
y2 = 10x
Solution:
The given equation of parabola is:
y2 = 10x, which is of the form y2 = 4ax.
Ex 10.2 Class 11 Maths Question 6.
x2 = -9y
Solution:
The given equation of parabola is:
x2 = -9y, which is of the form x2 = -4ay.
In each of the Exercises 7 to 12, find the equation of the parabola that satisfies the given conditions:
Ex 10.2 Class 11 Maths Question 7.
Focus (6, 0); directrix x = -6
Solution:
It is given that the focus (6, 0) lies on the x-axis, therefore, the x-axis is the axis of parabola. Also, the directrix is x = -6, i.e., x = -a and focus (6, 0), i.e., (a, 0). The equation of parabola is of the form y2 = 4ax.
Therefore, the required equation of the parabola is:
y2 = 4 × 6x ⇒ y2 = 24x
Ex 10.2 Class 11 Maths Question 8.
Focus (0, -3); directrix y = 3
Solution:
It is given that the focus (0, -3) lies on the y-axis, therefore, y-axis is the axis of parabola. Also, the directrix is y = 3, i.e., y = a and focus (0, -3), i.e., (0, -a). The equation of parabola is of the form x2 = -4ay.
Therefore, the required equation of the parabola is:
x2 = – 4 × 3y ⇒ x2 = -12y
Ex 10.2 Class 11 Maths Question 9.
Vertex (0, 0); focus (3, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (3, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis.
∴ The equation of the parabola is of the form y2 = 4ax.
The required equation of the parabola is:
y2 = 4 × 3x ⇒ y2 = 12x.
Ex 10.2 Class 11 Maths Question 10.
Vertex (0, 0); focus (-2, 0)
Solution:
Since the vertex of the parabola is at (0, 0) and focus is at (-2, 0).
∴ y = 0 ⇒ The axis of parabola is along x-axis.
∴ The equation of the parabola is of the form y2 = –4ax.
Therefore, the required equation of the parabola is:
y2 = –4 × 2x ⇒ y2 = -8x
Ex 10.2 Class 11 Maths Question 11.
Vertex (0, 0), passing through (2, 3) and axis is along x-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and the axis is along x-axis.
∴ The equation of the parabola is of the form y2 = 4ax.
Since the parabola passes through the point (2, 3).
Ex 10.2 Class 11 Maths Question 12.
Vertex (0, 0), passing through (5, 2) and symmetric with respect to y-axis.
Solution:
Since the vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
∴ The equation of the parabola is of the form x2 = 4ay.
Since the parabola passes through the point (5, 2).
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