JEE Main 2024 — Parabola Question with Solution

Line perpendicular to $$2x-y=10$$ have slope $$=\frac{-1}{2}$$

$$\Rightarrow$$ Line tangent to parabola $$y^2=4(x-9)$$ with slope $$m$$ is

$$\begin{array}{rl}& y=m(x-9)+\frac{1}{m},m=\frac{-1}{2}\\ & \Rightarrow y=\frac{-(x-9)}{2}-2\Rightarrow 2y=-x+9-4\\ & \Rightarrow 2y+x=5\end{array}$$

Solving the tangent and parabola we get point $$P$$

$$\begin{array}{rl}& {\left(\frac{5-x}{2}\right)}^2=4(x-9)\Rightarrow x^2-10x+25=16x-144\\ & \Rightarrow x^2-26x+169=0\Rightarrow (x-13{)}^2=0\\ & \Rightarrow P\equiv (13,-4)\end{array}$$

Distance of $$P$$ from the centre of circle $$(7,4)$$ is $$\sqrt{(13-7{)}^2+(-4-4{)}^2}=\sqrt{36+64}=10$$ units.