Question

Normals at $P , Q , R$ are drawn to $y^{2}=4 x$ which intersect at $(3,0)$. Then

Column I		Column II
A	Area of $\triangle PQR$	P	$2$
B	Radius of circumcircle of $\triangle PQR$	Q	$\frac{5}{2}$
C	Centroid of $\triangle PQR$	R	$\left(\frac{5}{2}, 0\right)$
D	Circumcentre of $\triangle PQR$	S	$\left(\frac{2}{3}, 0\right)$

• A. $( A ) \rightarrow( P ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( S ) ;( D ) \rightarrow( R )$
• B. $( A ) \rightarrow( Q ) ;( B ) \rightarrow( P ) ;( C ) \rightarrow( S ) ;( D ) \rightarrow( R )$
• C. $( A ) \rightarrow( R ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( S ) ;( D ) \rightarrow( P )$
• D. $( A ) \rightarrow( S ) ;( B ) \rightarrow( Q ) ;( C ) \rightarrow( P ) ;( D ) \rightarrow( R )$

Answer 1

Answer: (A)

It is given that
$y^{2}=4$ x
The equation of normal is
$y = mx -2 m -m^{3}(1)$
which is passing through the point $(3,0)$.
Therefore, Eq. (1) becomes
$0=3 m-2 m-m^{3}$
$m^{3}-m=0$
$ \Rightarrow m=0,-1,1$
That is, $\left(m_{1}^{2}-2 m_{1}\right),\left(m_{2}^{2},-2 m_{2}\right),\left(m_{3}^{2}-2 m_{3}\right)$
; hence, we have the points $P(0,0), Q(1,2)$ and $R(1,-2)$ as shown in the following figure:

(A) The area of triangle $P Q R$ is
$\frac{1}{2} \times 1 \times 4=2$
Therefore, (A) $\rightarrow$ (P).
(B) The radius of circumcircle is
$R=\frac{a b c}{4 \times \text { Area of } \Delta P Q R}=\frac{4 \sqrt{5} \sqrt{5}}{4 \times 2}=\frac{5}{2}$
Therefore, (B) $\rightarrow( Q )$.
(C) The centroid of $\triangle PQR$ is
$\left(\frac{0+1+1}{3}, \frac{0+2-2}{3}\right)=\left(\frac{2}{3}, 0\right)$
Therefore, (C) $\rightarrow$ (S).
(D) The circumradius of $\triangle PQR$ is $5 / 2$ and the circumcentre of $\Delta PQR$ is $\left(\frac{5}{2}, 0\right)$

Therefore, $(D) \to (R)$