Find the equation of the normal to curve
Question:

Find the equation of the normal to curve $y^{2}=4 x$ at the point $(1,2)$.

Solution:

The equation of the given curve is $y^{2}=4 x$.

Differentiating with respect to x, we have:

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

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

 

$\left.\therefore \frac{d y}{d x}\right]_{(1,2)}=\frac{2}{2}=1$

Now, the slope of the normal at point $(1,2)$ is $\left.\frac{\frac{-1}{d y}}{d x}\right]_{(1,2)}=\frac{-1}{1}=-1$.

∴Equation of the normal at (1, 2) is y − 2 = −1(x − 1).

$\Rightarrow y-2=-x+1$

 

$\Rightarrow x+y-3=0$

 

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