Find the equation of the tangent and normal to the given curve at the given points $y = x^{2} a t (0, 0)$

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Solution

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

On differentiating w.r.t. x, we get $\frac{d y}{d x} = 2 x$

$∴$ Slope of tangent at (0,0) is ${(\frac{d y}{d x})}_{(0, 0)} = 2 \times 0 = 0$

Hence, the slope of the tangent at (0,0) having slope 0 is

$y - 0 = 0 (x - 0) \Rightarrow y = 0$

the slope of normal at (0,0) is $\frac{- 1}{S l o p e o f t a n g e n t a t (0, 0)} = \frac{1}{0}$

Which is not defined.

Therefore, the equation of the normal at (x_0,y_0)=(0,0) is given by

$x = x_{0} = 0$

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