Find the length of the tangent and the sub-ta | vedclass

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(A) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.At th

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$2at\sqrt{t^2 + 1}, 2at^2$

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(A) Given the curve $y^2 = 4ax$. Differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4a$,which implies $\frac{dy}{dx} = \frac{2a}{y}$.At the point $(at^2, 2at)$,the slope of the tangent is $m = \frac{2a}{2at} = \frac{1}{t}$.The length of the tangent is given by $|y| \sqrt{1 + (\frac{dx}{dy})^2} = |2at| \sqrt{1 + t^2} = 2at\sqrt{t^2 + 1}$ (assuming $t > 0$).The length of the sub-tangent is given by $|y \frac{dx}{dy}| = |2at \cdot t| = 2at^2$.

Find the length of the tangent and the sub-tangent for the curve $y^2 = 4ax$ at the point $(at^2, 2at)$.

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