For the curve y^2 = 4ax,the lengths of the 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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$2at\sqrt{t^2 + 1}, 2at^2, 2a\sqrt{t^2 + 1}, 2a$

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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 $m = \frac{dy}{dx} = \frac{2a}{2at} = \frac{1}{t}$.$1$. Length of Subtangent = $|\frac{y}{m}| = |\frac{2at}{1/t}| = 2at^2$.$2$. Length of Tangent = $|y \sqrt{1 + \frac{1}{m^2}}| = |2at \sqrt{1 + t^2}| = 2at\sqrt{t^2 + 1}$.$3$. Length of Subnormal = $|y \cdot m| = |2at \cdot \frac{1}{t}| = 2a$.$4$. Length of Normal = $|y \sqrt{1 + m^2}| = |2at \sqrt{1 + \frac{1}{t^2}}| = |2at \frac{\sqrt{t^2 + 1}}{t}| = 2a\sqrt{t^2 + 1}$.Thus,the lengths are $2at\sqrt{t^2 + 1}, 2at^2, 2a\sqrt{t^2 + 1}, 2a$.

For the curve $y^2 = 4ax$,the lengths of the tangent,subtangent,normal,and subnormal at the point $(at^2, 2at)$ are respectively:

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