Question

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

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

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

Answer 1

Answer: (A)

It is given that
$y^2=4$ x
The equation of normal is
$y=mx-2m-m^3(1)$
which is passing through the point $(3,0)$.
Therefore, Eq. (1) becomes
$0=3m-2m-m^3$
$m^3-m=0$
$\Rightarrow m=0,-1,1$
That is, $\left(m_1^2-2m_1\right),\left(m_2^2,-2m_2\right),\left(m_3^2-2m_3\right)$
; 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 $PQR$ is
$\frac{1}{2}\times 1\times 4=2$
Therefore, (A) $\to$ (P).
(B) The radius of circumcircle is
$R=\frac{abc}{4\times \text{ Area of }\Delta PQR}=\frac{4\sqrt{5}\sqrt{5}}{4\times 2}=\frac{5}{2}$
Therefore, (B) $\to (Q)$.
(C) The centroid of $△PQR$ is
$\left(\frac{0+1+1}{3},\frac{0+2-2}{3}\right)=\left(\frac{2}{3},0\right)$
Therefore, (C) $\to$ (S).
(D) The circumradius of $△PQR$ is $5/2$ and the circumcentre of $\Delta PQR$ is $\left(\frac{5}{2},0\right)$

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