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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class11  >>  Coordinate Geometry

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

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

1 Answer

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.
answered Feb 6, 2014 by sreemathi.v
edited Sep 23, 2014 by sharmaaparna1
 

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