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Q. For the parabola $y^2 = 4x$, the point $P$ whose focal distance is $17$, is

KCETKCET 2009Conic Sections

Solution:

Given, $y^{2}=4 x$
Let $P(h, k)$ be any point on the parabola
$\therefore \,\,\,\,\,\,\,(h-1)^{2}+(k-0)^{2}=17^{2}$
Also, $k^{2}=4 h$
$\therefore \,\,\,\,\,\,\,h^{2}+1-2 h+4 h=289$
$\Rightarrow \,\,\,\,\,\,\, h^{2}+2 h-288=0$
$\Rightarrow \,\,\,\,\,\,\,(h+18)(h-16)=0$
$\Rightarrow \,\,\,\,\,\,\, h=16 \,\,\,\,\,\,\,(\because h$ cannot be negative)
$\therefore \,\,\,\,\,\,\, k^{2}=64$
$\Rightarrow \,\,\,\,\,\,\, k=\pm 8$
$\therefore $ Points are $(16,8)$ or $(16,-8)$