Let S be the focus of the parabola y^2=4ax an | vedclass

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(C) The focus of the parabola $y^2=4ax$ is $S(a, 0)$.Let the coordinates of $P$ be $(at^2, 2at)$ and $Q$ be $(\frac{a}{t^2}, -\frac{2a}{t})$.Since $SP

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$\frac{1}{a}$

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(C) The focus of the parabola $y^2=4ax$ is $S(a, 0)$.Let the coordinates of $P$ be $(at^2, 2at)$ and $Q$ be $(\frac{a}{t^2}, -\frac{2a}{t})$.Since $SP = \alpha$,the distance from $S(a, 0)$ to $P(at^2, 2at)$ is $\alpha = \sqrt{(at^2-a)^2 + (2at-0)^2} = \sqrt{a^2(t^2-1)^2 + 4a^2t^2} = \sqrt{a^2(t^4-2t^2+1+4t^2)} = \sqrt{a^2(t^2+1)^2} = a(t^2+1)$.Similarly,for $SQ = \alpha^{\prime}$,the distance from $S(a, 0)$ to $Q(\frac{a}{t^2}, -\frac{2a}{t})$ is $\alpha^{\prime} = \sqrt{(\frac{a}{t^2}-a)^2 + (-\frac{2a}{t}-0)^2} = \sqrt{a^2(\frac{1-t^2}{t^2})^2 + \frac{4a^2}{t^2}} = \sqrt{\frac{a^2(1-2t^2+t^4+4t^2)}{t^4}} = \sqrt{\frac{a^2(1+t^2)^2}{t^4}} = \frac{a(1+t^2)}{t^2}$.Now,$\frac{1}{\alpha} + \frac{1}{\alpha^{\prime}} = \frac{1}{a(t^2+1)} + \frac{t^2}{a(1+t^2)} = \frac{1+t^2}{a(1+t^2)} = \frac{1}{a}$.

Let $S$ be the focus of the parabola $y^2=4ax$ and $PQ$ be a focal chord such that $SP=\alpha$ and $SQ=\alpha^{\prime}$. Then $\frac{1}{\alpha}+\frac{1}{\alpha^{\prime}}=$

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