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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of normal for the parabola $y^2=4ax$ in parametric form

$\begin{array}{1 1}(a)\;y+x=2at+at^3\\(b)\;y^2+x^2=2at+at^3\\(c)\;y+x=at+at^2\\(d)\;y+2x=at+at^3\end{array}$

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Given equation of the parabola is $y^2=4ax$
Differentiating w.r.to $x$, we get $2y\large\frac{dy}{dx}$$=4a$
Slope of tangent to this curve at $ (x_1,y_1)$ is $\large\frac{dy}{dx}=\frac{2a}{y_1}$
Slope of normal$=\large\frac{-1}{dy/dx}$$=-\frac{y_1}{2a}$
$\therefore$ For the parabola $y^2=4ax$ equation of normal at $(x_1,y_1)$ is
Parametric coordinates of the parabola is given by $(x_1,y_1)=(at^2,2at)$
$\therefore$ Eqn. of normal becomes $(y-2at)=-\large\frac{2at}{2a}$$(x-at^2)$
Hence (a) is the correct answer.
answered Feb 5, 2014 by sreemathi.v
edited Mar 27, 2014 by rvidyagovindarajan_1

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