Ask Questions, Get Answers

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

Find the equation of tangent to the parabola $(y-6)^2=4a(x-p)$ at $(at^2,2at)$?

$\begin{array}{1 1}(a)\;t(y+b)=x-p+at^2\\(b)\;t(y-b)=x-p+at^2\\(c)\;t(y+b)=x+p+at^2\\(d)\;t(y+b)=x+p-at^2\end{array}$

Can you answer this question?

1 Answer

0 votes
We know that the equation of tangent in parametric form is
$ty=x+at^2$(for parabola $y^2=4ax$)
Now converting the equation by
$(y-6)^2=4a(x-p)$ in the form of $y^2=4ax$
Hence replacing
Now equation of tangent becomes
This is the required equation of tangent.
Hence (b) is the correct answer.
answered Feb 6, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1

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