Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

The normal to the parabola $y^2=8x$ at the point $(2,4)$ meets the parabola again at the point.

$\begin{array}{1 1}(A)\;(-18,-12) \\(B)\;(-18,12) \\(C)\;(18,12) \\(D)\;(18,-12) \end{array}$

Can you answer this question?

1 Answer

0 votes
$y^2=8x$ -----(1)
=> $ 2y. \large\frac{dy}{dx}$$=8$
Slope of tangent at $(2,4)= \bigg[\large\frac{dy}{dx}\bigg]_{(2,4)}$
Slope of normal at $(2.4)=-1$
Equation of normal at $(2,4)$ is
$x+y-6=0$ ----(2)
Solving (1) and (2) we get,
and $x=18,y=-12$
$\therefore $ other point is $(18,-12)$
Hence D is the correct answer.
answered Apr 9, 2014 by meena.p

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App