Question

$\begin{array}{cc}Column1& Column2\\ \text{a. Tangents are drawn from the point}(2,3)& \text{p.}(9,-6)\\ \text{to the parabola}{y}^2=4x\text{then points of contact are}& \\ \text{b. From a point}P\text{on the circle}{x}^2+y^2=5,& \\ \text{the equation of chord of contact to the parabola}& \\ y^2=4x\text{is}y=2(x-2),\text{then the coordinate of}& \\ \text{point}P\text{will be}& \text{q.}(1,2)\\ \text{c.}P(4,-4),Q\text{are points on parabola}{y}^2=4x& \\ \text{such that area of}\Delta POQ\text{is 6 sq. units where}& \\ O\text{is the vertex, then coordinates of}Q\text{may be}& \text{r.}(-2,1)\\ \text{d. The chord of contact w.r.t any point on the}& \\ \text{directrix of the parabola}(y-2{)}^2=4x\text{passes}& \\ \text{through the point}& \text{s.}(4,4)\end{array}$

Which of the following is correct option

• A. $a\to q,r,b\to q,c\to p,q,d\to s$
• B. $a\to q,s,b\to r,c\to p,qd\to q$
• C. $a\to q,b\to r,c\to p,d\to q$
• D. $a\to q,s,b\to q,c\to p,s,d\to s$

Answer 1

The correct option is B $a\to q,s,b\to r,c\to p,qd\to q$

$\left(a\right).$
Let the point on the parabola be
$(t^2,2t)$
The equation of the tangent will be,
$2ty=2(x+t^2)$
$\Rightarrow ty=x+t^2$
It passes through the point (2, 3),
$3t=2+t^2\Rightarrow t=1,2$
The point of contact will be
$(1,2)$ or $(4,4)$
$\left(a\right)\to \left(q\right),\left(s\right)$

$\left(b\right).$
Let a point on the circle be $P(x_1,y_1)$
Then the chord of contact of the parabola w.r.t $P$ is
$y{y}_1=2(x+x_1)$
Comparing with $y=2(x-2)$,
We have $y_1=1$ and $x_1=-2$,
Which also satisfy the circle.
$\left(a\right)\to \left(r\right)$

$\left(c\right).$
Point $Q$ on the parabola is at $(t^2,2t)$
Now, the area of triangle $OPQ$ is
$\left|\frac{1}{2}\left|\begin{array}{cc}0& 0\\ 4& -4\\ t^2& 2t\\ 0& 0\end{array}\right|\right|=6\Rightarrow 8t+4t^2=\pm 12\Rightarrow t^2+2t+1=1\pm 3\Rightarrow (t+1{)}^2=-2\text{or}4\Rightarrow (t+1{)}^2=4\Rightarrow t=-1\pm 2\Rightarrow t=-3,1$
Hence, the point $Q$ is $(1,2)\text{or}(9,-6)$.
$\left(c\right)\to \left(p\right),\left(q\right)$

$\left(d\right).$
Points (1, 2) and (-2, 1) satisfy both the curves.