Let $L_1$ be the length of the common chord of the curves $x^2+y^2=9$ and $y^2=8 x$, and $L_2$ be the length of the latus rectum of $y^2=8 x$, then:

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Aicharya · Accepted Answer

We have : $x^2+(8 x)=9$ $\Rightarrow x^2+9 x-x-9=0$ $\Rightarrow x(x+9)-1(x+9)=0$ $\Rightarrow(x+9)(x-1)=0 \quad \Rightarrow x=-9,1$ for $x=1, y= \pm 2 \sqrt{2 x}= \pm 2 \sqrt{2}$ $\sqrt{(2 \sqrt{2}+2 \sqrt{2})^2+(1-1)^2}=4 \sqrt{2}$ $\mathrm{L}_n=$ Length of latus rectum $=4 a=4 	imes 2=8$ $\mathrm{L}_1 \lt \mathrm{L}_2$

Let $L_1$ be the length of the common chord of the curves $x^2+y^2=9$ and $y^2=8 x$, and $L_2$ be the length…

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