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Find the equation of the normal to the curve $x^2=4y$ which passes through the point $\left ( -1,4 \right ).$

Answers (1)

Suppose the normal at $P(x_1,y_1)$ on the parabola $x^2=4y$ passes through $\left ( -1,4 \right )$

$\because P(x_1,y_1)\: \: lies \: \: on\: \: x^2=4y$

$\therefore {x_{1}}^{2}=4y_1\: \: \: \: -(i)$

The equation of the curve is $x^2=4y$

Differentiative w.r.t x, we have

$2x=4\frac{\mathrm{d} y}{\mathrm{d} x}$

$\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\frac{x}{2}\Rightarrow \left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )_{x_1,y_1}=\frac{x_1}{2}$

The equation $(x_1,y_1)$ of normal at $P(x_1,y_1)$ is

$y-y_1=\frac{-1}{\frac{\mathrm{d} y}{\mathrm{d} x}}\left ( x-x_1 \right )$

$y-y_1=\frac{-2}{x_1}\left ( x-x_1 \right )\: \: \: \: \: -(ii)$

$\because It\: \: \: passes\: \: \: through\; \; (-1,4)$

$\therefore putting\: \: \: x=-1\: \: and\: \: y=4,we\: get$

$\Rightarrow 4-y_1=\frac{-2}{x_1}(-1-x_1)\Rightarrow 4-y_1=\frac{2}{x_1}(1+x_1)$ $\Rightarrow 4-y_1=\frac{2}{x_1}(1+x_1)\Rightarrow 4x_1-x_1y_1=2+2x_1$

$\Rightarrow 2x_1=2+x_1y_1\Rightarrow \frac{2x_1-2}{x_1}=y_1\: \: \: \: (iii)$

Eliminating $y_1$ from equation (i), we have

${x_{1}}^{2}=4\left ( \frac{2x_1-2}{x_1} \right )$

${x_{1}}^{3}=8x_1-8\Rightarrow x_1=2$

Putting $x_1=2\: \: in\: \: (iii),we\: \: get\: \: \: y_1=1$

Putting values of $x_1,y_1,$ in (ii), we get

$(y-1)=-1(x-2)$

$x+y-3=0$

This is the required equation of normal to the given curve.

Posted by

Ravindra Pindel

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