Let $\mathrm{L}_1$ 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 $ \begin{aligned} &x^2+(8 x)=9 \\ &x^2+9 x-x-9=0 \\ &x(x+9)-1(x+9)=0 \\ &(x+9)(x-1)=0 \\ &x=-9,1 \\ &\text { for } x=1, y=\pm 2 \sqrt{2 x}=\pm 2 \sqrt{2} \\ &\mathrm{~L}_1=\text { Length of } \mathrm{AB} \\ &\quad=\sqrt{(2 \sqrt{2}+2 \sqrt{2})^2+(1-1)^2}=4 \sqrt{2} \\ &\mathrm{~L}_2=\text { Length of latus rectum } \\ &=4 a=4 \times 2=8 \\ &\mathrm{~L}_1 < \mathrm{L}_2 \end{aligned} $

Let $\mathrm{L}_1$ be the length of the common chord of the curves $x^2+y^2=9$ and $y^2=8 x$, and $L_2$ be…

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