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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

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

$y-6=Y$

$x-p=X$

$Y^2=4aX$

Now equation of tangent becomes

$tY=X+at^2$

$t(y-6)=x-p+at^2$

This is the required equation of tangent.

Hence (b) is the correct answer.

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...