If A(at^2, 2at),B(a/t^2, -2a/t),and C(a, 0),t | vedclass

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(C) Given points are $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$.$CA = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = a\sqrt

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$H.M.$ of $CA$ and $CB$

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(C) Given points are $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$.$CA = \sqrt{(at^2 - a)^2 + (2at - 0)^2} = \sqrt{a^2(t^2 - 1)^2 + 4a^2t^2} = a\sqrt{t^4 - 2t^2 + 1 + 4t^2} = a\sqrt{(t^2 + 1)^2} = a(t^2 + 1)$.$CB = \sqrt{(\frac{a}{t^2} - a)^2 + (-\frac{2a}{t} - 0)^2} = \sqrt{a^2(\frac{1 - t^2}{t^2})^2 + \frac{4a^2}{t^2}} = a\sqrt{\frac{1 - 2t^2 + t^4 + 4t^2}{t^4}} = a\sqrt{\frac{(1 + t^2)^2}{t^4}} = a\frac{1 + t^2}{t^2}$.$H.M.$ of $CA$ and $CB$ is given by $\frac{2(CA)(CB)}{CA + CB}$.$H.M. = \frac{2 \cdot a(t^2 + 1) \cdot a(\frac{1 + t^2}{t^2})}{a(t^2 + 1) + a(\frac{1 + t^2}{t^2})} = \frac{2a^2(t^2 + 1)^2 / t^2}{a(t^2 + 1)(1 + 1/t^2)} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)(\frac{t^2 + 1}{t^2})} = \frac{2a(t^2 + 1)^2 / t^2}{(t^2 + 1)^2 / t^2} = 2a$.Thus,$2a$ is the $H.M.$ of $CA$ and $CB$.

If $A(at^2, 2at)$,$B(a/t^2, -2a/t)$,and $C(a, 0)$,then $2a$ is equal to

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