Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Let the normal at the end points of a variable chord of the parabola <img alt="\mathrm{y^2-8 y+

Let the normal at the end points of a variable chord \mathrm{A B} of the parabola \mathrm{y^2-8 y+2 x=0} are perpendicular. If the tangents at \mathrm{A \: and \: B} will intersect at \mathrm{a x+b=0} then value of \mathrm{|2 a+b|} equals

Option: 1

13


Option: 2

10


Option: 3

9


Option: 4

15


Answers (1)

best_answer

As the normals are \perp therefore tangents will also be perpendicular and the \perp tangents always intersect each other at directrix.
Now, equation of parabola is
\mathrm{ y^2-8 y+2 x=0 }

\mathrm{ \Rightarrow(y-4)^2=-2 x+16 \Rightarrow(y-4)^2=-2(x-8) }

\mathrm{ Y^2=4\left(-\frac{1}{2}\right) X, X=x-8, Y=y-4 }

\mathrm{ \therefore \quad \text{ Equation of directrix is} }

\mathrm{ X=\frac{1}{2} \text { i.e., } x-8=\frac{1}{2} \text { or } 2 x-17=0 }

\mathrm{ Now \: a x+b=2 x-17 }

\mathrm{ \therefore|2 a+b|=13 }



 

Posted by

Rakesh

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