Mathematics · Parabola

JEE Main 2024 — 5 April, Shift 2 — Question 29

Let a line perpendicular to the line $2 \mathrm{x}-\mathrm{y}=10$ touch the parabola $y^{2}=4(x-9)$ at the point $P$ . The distance of the

point P from the centre of the circle $x^{2}+y^{2}-14 x-8 y+56=0$ is $\qquad$

Answer: 10

Numerical answer — enter this value.

$y^{2}=4(x-9)$

slope of tangent $=\frac{-1}{2}$

Point of contact $\mathrm{P}\left(9+\frac{1}{\left(-\frac{1}{2}\right)^{2}}, \frac{2 \times 1}{\frac{-1}{2}}\right)$

$\mathrm{P}(13,-4)$

center of circle $C(7,4)$

distance $\mathrm{CP}=\sqrt{(13-7)^{2}+(-4-4)^{2}}$

$=10$

Answer key and solution verified before publishing.

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.