Find the equations of the tangent and normal | vedclass

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Question

(A) The equation of the given parabola is $y^{2}=4ax$. On differentiating $y^{2}=4ax$ with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies

Vedclass · Accepted Answer

Tangent: $ty = x + at^{2}$,Normal: $y = -tx + 2at + at^{3}$

Vedclass · Answer

(A) The equation of the given parabola is $y^{2}=4ax$. On differentiating $y^{2}=4ax$ with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$. The slope of the tangent at $(at^{2}, 2at)$ is $\left. \frac{dy}{dx} \right|_{(at^{2}, 2at)} = \frac{2a}{2at} = \frac{1}{t}$. The equation of the tangent is $y - 2at = \frac{1}{t}(x - at^{2})$,which simplifies to $ty - 2at^{2} = x - at^{2}$,or $ty = x + at^{2}$. The slope of the normal is the negative reciprocal of the tangent slope,which is $-t$. The equation of the normal is $y - 2at = -t(x - at^{2})$,which simplifies to $y - 2at = -tx + at^{3}$,or $y = -tx + 2at + at^{3}$.

Find the equations of the tangent and normal to the parabola $y^{2}=4ax$ at the point $(at^{2}, 2at)$.

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