Find the equations of the tangent and the normal to the curve `y=x^2` at `(0,\ 0)` at the indicated points

Answer

Equation of tangent is: `y=0`

Equation of normal is: `x=0`

Explanation

ATQ,

`y=x^{2}`

Differentiating both sides w.r.t. `x` ,

`\frac{dy}{dx} = 2x`


Also given `(x_{1}, y_{1})=(0,0)`

Slope of tangent, `m=(\frac{d y}{d x})_{(0,0)}=2(0)=0`


Equation of tangent is ,
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