Ex 11.2, 12 - Find parabola: Vertex (0, 0) passing (5, 2)

Ex 11.2,  12 - Chapter 11 Class 11 Conic Sections - Part 2
Ex 11.2,  12 - Chapter 11 Class 11 Conic Sections - Part 3

This video is only available for Teachoo black users

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 10.2, 12 Find the equation of the parabola that satisfies the following conditions: Vertex (0, 0) passing through (5, 2) and axis is along y-axis Given that axis is along the y-axis, So, equation of parabola is of the form x2 = 4ay or x2 = −4ay Plotting point (5, 2) Since point lie of the 1st quadrant & parabola passes through the point (5, 2) So, the parabola is of the form Hence equation of parabola is x2 = 4ay Point (5, 2) will satisfy the equation of parabola Putting x = 5 & y = 2 in (1) (5)2 = 4a(2) 25 = 8a 8a = 25 a = 𝟐𝟓/𝟖 Hence equation of parabola is x2 = 4ay Putting value of a = 25/8 x2 = 4(25/8)y x2 = 𝟐𝟓/𝟐 𝐲

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.