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 $y^2=4 x, A B$ and $C D$ 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
$(x-1)^2=a y+b$
It passing $(0,1) \&(3,3)$
so $1= a + b$...(1)
$4=3 a+b$...(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 $x^2+y^2+2 g x+2 f y+c=0$
fourth point $M\left(t_5\right)$ putting the value $\left(t^2, 2 t\right)$ 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 $\left(a^2, 2 a t\right) \Rightarrow(4,-4)$