Exam

JEE Main

Year

2025

Date

23 January

Shift

Evening Shift

Subject

Mathematics

Chapter

Parabola

JEE Main 2025 · 23 January · Evening Shift · Mathematics · Parabola

Let the shortest distance from (a, 0), a 0 , to the parabola y^2=4...

Let the shortest distance from $(\mathrm{a}, 0), \mathrm{a}\gt0$, to the parabola $y^2=4 x$ be 4 . Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is :

Step-by-step solution

Normal at P $$\begin{aligned} & y+ t x=2 t+t^3 \\ & \uparrow \\ &(\mathrm{a}, 0) \\ & \mathrm{at}= 2 \mathrm{t}+\mathrm{t}^3 \\ & \mathrm{a}= 2+\mathrm{t}^2 \\ & \mathbb{R}\left(2+\mathrm{t}^2, 0\right) \end{aligned}$$$$\begin{aligned} & \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16 \\ & 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \\ & \mathrm{a}=5 \mathbb{R}(5,0) \end{aligned}$$$Focus$(1,0) (1,0) \&(5,0)$will be tha end pts. of diameter$\Rightarrow \mathrm{Eg}^{\mathrm{n}}$of circle is$$\begin{aligned} & (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0 \end{aligned}$$