Let the shortest distance from $(\mathrm{a}, 0), \mathrm{a} \gt 0$, 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 :

Question

Aicharya · Accepted Answer

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
$\Rightarrow$ Eqn of circle is
$\begin{aligned}
& (x-1)(x-5)+y^2=0 \\ & x^2+y^2-6 x+5=0
\end{aligned}$ /

Let the shortest distance from $(\mathrm{a}, 0), \mathrm{a} \gt 0$, to the parabola $y^2=4 x$ be 4. Then the…

Solution