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NCERT · Class 11 · Maths

Chapter 10: Conic Sections

Practice each question, then open the answer or worked solution.

Exercise 10.2

12 questions
Q11Written answer

Find the equation of the parabola whose vertex is (0, 0), passing through (2, 3) and axis is x-axis.

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Worked solution

Solution:

Since the axis is the x-axis, the equation is

y² = 4ax

Substitute point (2, 3):

3² = 4a × 2

9 = 8a

a = 9/8

Equation:

y² = 4 × (9/8)x

y² = (9/2)x

Multiplying by 2:

2y² = 9x

Answer: 2y² = 9x

Q12Written answer

Find the equation of the parabola whose vertex is (0, 0), passes through (5, 2) and is symmetric about y-axis.

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Worked solution

Solution: 

A parabola symmetric about the y-axis has the equation:

x² = 4ay

Substitute point (5, 2):

25 = 4a × 2

25 = 8a

a = 25/8

Equation:

x² = 4 × (25/8)y

x² = (25/2)y

Multiplying by 2:

2x² = 25y

Answer: 2x² = 25y