Mathematics - Parabola Question with Solution | TestHub

$y^2=kx-8$ $y^2=k(x-8/K)$ $r(8/K,0);4a=K$ $a=\frac{K}{4}$ We know that $rz=a$ $\Rightarrow \left|\frac{8}{K}-1\right|=\frac{K}{4}$ $\Rightarrow \frac{8}{K}-\frac{K}{4}=1$ $\Rightarrow 32-K^2=4K$ $\Rightarrow K^2+4K-32=0$ $\Rightarrow K(K+8)-4(K+8)=0$ $\mathrm{K}=4$ or $\mathrm{K}=-8\mathrm{X}$