Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Find the equation of the chord of the parabola x2 = 8y which is bisected at (3, 2)Option: 1 <img alt="4 x+3 y+1=0" src="https://learn.care

Find the equation of the chord of the parabola x2 = 8y which is bisected at (3, 2)

Option: 1

4 x+3 y+1=0


Option: 2

4 x-3 y+1=0


Option: 3

4 y+3 x+1=0


Option: 4

4 y-3 x+1=0


Answers (1)

best_answer

The equation of the chord of parabola ,whose midpoint P(x1,y1) is T = S_1

So, the equation of the chord of the parabola x2 = 8y, which is bisected at (3, 2) is

\\\begin{aligned} & T=S_{1} \\ \Rightarrow & xx_{1}-2 a\left(y+y_{1}\right)=x_{1}^{2}-4 a y_{1} \\ \Rightarrow & 3 x-4(y+2)=9-8.2=-7 \\ \Rightarrow & 3 x-4 y-1=0 \\ \Rightarrow & 4 y-3 x+1=0 \end{aligned}

Posted by

jitender.kumar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Latest Question