Q72 / 180 · KCET 2009 · Mathematics
For the parabola $\mathrm{y}^{2}=4 \mathrm{x}$, the point $P$ whose focal distance is 17 , is
- A$(8,8)$ or $(8,-8)$
- B$(4,8)$ or $(4,-8)$
- C$(2,8)$ or $(2,-8)$
- Correct answerD$(16,8)$ or $(16,-8)$
Solution
Given, $\mathrm{y}^{2}=4 \mathrm{x}$
Let $\mathrm{P}(\mathrm{h}, \mathrm{k})$ be any point on the parabola
$\therefore \quad(\mathrm{h}-1)^{2}+(\mathrm{k}-0)^{2}=17^{2}$
Also, $\mathrm{k}^{2}=4 \mathrm{~h}$
$\therefore \quad \mathrm{h}^{2}+1-2 \mathrm{~h}+4 \mathrm{~h}=289$
$\Rightarrow \quad \mathrm{h}^{2}+2 \mathrm{~h}-288=0$
$\Rightarrow \quad(\mathrm{h}+18)(\mathrm{h}-16)=0$
$\Rightarrow \quad \mathrm{h}=16 \quad(\because h$ cannot be negative $)$
$\therefore \quad \mathrm{k}^{2}=64$
$\Rightarrow \quad \mathrm{k}=\pm 8$
$\therefore$ Points are $(16,8)$ or $(16,-8)$.
Hence, the correct answer is option: 'D'.