Example 8 - Chapter 10 Class 11 Conic Sections

Last updated at May 6, 2024 by

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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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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo