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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equations of the tangent and normal to the parabola \(y^2 = 4ax\) at the point \((at^2, 2at).\)

$\begin{array}{1 1} (A)\;x-ty+at^2=0;xt+y-at(2+t)=0 \\ (B)\;x-ty-at^2=0;xt+y-at(2+t)=0 \\(C)\;x-ty+at^2=0;xt-y+at(2+t)=0 \\(D)\;x-ty+at^2=0;xt+y+at(2+t)=0 \end{array} $

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Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
The equation of the given curve is $y^2=4ax$
Differentiating w.r.t $x$ we get,
$2y\large\frac{dy}{dx}$$=4a$
$\large\frac{dy}{dx}=\frac{4a}{2y}=\frac{2a}{y}$
Hence the slope of the tangent to the curve is $\large\frac{dy}{dx}=\frac{2a}{y}$
Slope of the tangent to the curve at $(at^2,2at)$
$\large\frac{dy}{dx}_{(at^2,2at)}=\frac{2a}{2at}$
$\qquad\qquad\;=\large\frac{1}{t}$
Step 2:
Therefore equation of the tangent is $(y-2at)=\large\frac{1}{t}$$(x-at^2)$
On simplifying we get,
$x-ty+at^2=0$
Slope of the normal to the curve at $(at^2,2at)$ is $-t$
Therefore equation of the normal is $(y-2at)=-t(x-at^2)$
$\Rightarrow xt+y-2at-at^2=0$
$\Rightarrow xt+y-at(2+t)=0$
answered Jul 12, 2013 by sreemathi.v
 

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