Let the shortest distance from (a, 0), a0, to the parabola y^2=4 x be 4 .

Question

Let the shortest distance from $(a, 0), a>0$, 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 :

Accepted Answer

$$\begin{aligned}
&\text { Normal at } \mathrm{P}\
&\begin{gathered}
y+t x=2 t+t^3 \
\uparrow \
(a, 0)
\end{gathered}
\end{aligned}$$

$$\begin{aligned}
& \mathrm{at}=2 \mathrm{t}+\mathrm{t}^3 \
& \mathrm{a}=2+\mathrm{t}^2 \
& \mathrm{R}\left(2+\mathrm{t}^2, 0\right) \
& \mathrm{PR}=4 \Rightarrow 4+4 \mathrm{t}^2=16
\end{aligned}$$

$$\begin{aligned}
& \quad 4 \mathrm{t}^2=12 \Rightarrow \mathrm{t}^2=3 \
& \mathrm{a}=5, \mathrm{R}(5,0)
\end{aligned}$$

Focus $(1,0)$

$(1,0) \&(5,0)$ will be the end points of diameter

$\Rightarrow E q^{\mathrm{n}}$ of circle is

$$\begin{aligned}
& (x-1)(x-5)+y^2=0 \
& x^2+y^2-6 x+5=0
\end{aligned}$$

Solutions · 1