Find the equations of the tangent and normal | vedclass

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(A) The equation of the curve is $y=x^{2}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = 2x$.The slope of the tangent at $(0,0)$ is $

Vedclass · Accepted Answer

Tangent: $y=0$,Normal: $x=0$

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(A) The equation of the curve is $y=x^{2}$.On differentiating with respect to $x$,we get:$\frac{dy}{dx} = 2x$.The slope of the tangent at $(0,0)$ is $\left. \frac{dy}{dx} ight|_{(0,0)} = 2(0) = 0$.The equation of the tangent at $(0,0)$ is $y - 0 = 0(x - 0)$,which simplifies to $y = 0$.The slope of the normal at $(0,0)$ is $\frac{-1}{	ext{slope of tangent}} = \frac{-1}{0}$,which is undefined.For a vertical line passing through $(x_0, y_0) = (0,0)$,the equation is $x = x_0$,so the equation of the normal is $x = 0$.

Find the equations of the tangent and normal to the given curve $y=x^{2}$ at the point $(0,0)$.

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