Example 8 - Chapter 10 Class 11 Conic Sections

Last updated at August 2, 2026 by

Example 8 - Find equation of parabola symmetric about y-axis - Parabola - Beam problem

part 2 - Example 8 - Parabola - Beam problem - Concept wise - Chapter 10 Class 11 Conic Sections

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Example 8 Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3). Since the parabola is symmetric about y-axis Equation is of the form x2 = 4ay or x2 = – 4ay Now plotting point (2, −3) on graph Since (2, −3) lies in fourth quadrant So the parabola will be of the form Hence we use equation x2 = – 4ay. Now parabola passes through (2, −3) Putting x = 2, y = −3 in equation 22 = −4 × a × −3 4 = 12a 4/12 = a 1/3 = a a = 𝟏/𝟑 Equation of parabola is x2 = −4ay x2 = − 𝟒/𝟑 y

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