Question

The line $y=5-2x$ intersects the curve $3y^2=8x^2-6$ at two points, P and Q. Find the length of PQ and the midpoint of PQ.

Answer 1

Put $y=5-2x$ in $3y^2=8x^2-6$

$\Rightarrow 3(5-2x{)}^2=8x^2-6\Rightarrow 8x^2-6-3(4x^2-2x+25)=0$

$\Rightarrow 8x^2-6-12x^2+60x-75=0$

$\Rightarrow 4x^2-60x+81=0$

$\Rightarrow 4x^2-6x-54x+81=0\Rightarrow x=13-5,1-5$

$\therefore y=-22,3$ here pt. are $(1-5,3)$ & $(\frac{27}{2},-22)$

$\therefore distance=\sqrt{(13-5)-(-5{)}^2+(-22-3{)}^2}=\sqrt{850}$

mid pt. $=\left(\frac{13.5+1-5_7}{2}\frac{-22+3}{2}\right)=(7.5,\frac{-19}{2})$