Question

For the parabola $y^2=4x$, $AB$ and $CD$ are any two parallel chords having slope $1$. $C_1$ is a circle passing through $O$, $A$ and $B$ and $C_2$ is a circle passing through $O$, $C$ and $D$, where $O$ is origin. $C_1$ and $C_2$ intersect at

• A. $(4,-4)$
• B. $(-4,4)$
• C. $(4,4)$
• D. $(-4,-4)$

Answer 1

The correct option is A $(4,-4)$

Axis of parabola is bisector of parallel chords

$AB$ and $CD$ are parallel chords.

So axis is $x=1$
Equation of parabola is,
$(x-1{)}^2=ay+b$
It passes through $(0,1)$ & $(3,3)$
So, $1=a+b$ ...(1)
$4=3a+b$ ...(2)
From (1) & (2)
$a=\frac{3}{2}$ & $b=-\frac{1}{2}$
${(x-1)}^2=\frac{3}{2}(y-\frac{1}{3})$
Let points on $y^2=4ax$ are $A\left(t_1\right),B\left(t_1\right),C\left(t_1\right)$ and $D\left(t_1\right)$
So $t_1+t_2=2=t_3+t_4$
Equation of circle passing through $△OAB$ is
$x^2+y^2+2gx+2fy+c=0$
fourth point $M\left(t_5\right)$ puttimg the value $(t^2,2t)$ in circle we get four degree equation. In this equation
$t_1+t_2+t_5+0=0\Rightarrow t_6=-2$
Similarly circle passing through OCD & foyurth point $N\left(t_5\right)$ we have $t_1+t_2+t_5+0=0\Rightarrow t_5=-2$
It mean both point M and N are same
so common point $(a{t}^2,2at)\Rightarrow (4,-4)$