Question

Observe the following facts for a parabola :
(i) Axis of the parabola is the only line which can be the perpendicular bisector of the two chords of the parabola.
(ii) If $AB$ and $CD$ are two parallel chords of the parabola and the normals at $A$ and $B$ intersect at $P$ and the normals at $C$ and $D$ intersect at $Q$, then $PQ$ is a normal to the parabola.
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

Answer: (A)

Axis of parabola is bisector of parallel chord A B $\&CD$ are parallel chord.
so axis $x=1$

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