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

Question

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

Accepted Answer

Substituting $y^2=8x$ into $x^2+y^2=9$: $x^2+8x-9=0\Rightarrow(x+9)(x-1)=0\Rightarrow x=1$ (the valid root, $x=-9$ rejected).

At $x=1$, $y=\pm2\sqrt2$, so $L_1=4\sqrt2\approx5.66$. Latus rectum of $y^2=8x$: $L_2=8$.

Since $4\sqrt2<8$, $L_1<L_2$.

Solutions · 1