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.  

 

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