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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
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

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