Question

Match the following:

$\begin{array}{cccc}& \text{Column-I}& & \text{Column-II}\\ \left(a\right)& \text{The reflection of the point (t - 1, 2t + 2) in a line}& \left(p\right)& 5\\ & \text{is (2t + 1, t) then the line has slope equal to}& & \\ \left(b\right)& \text{If}\theta \text{is the angle between two tangents which}& \left(q\right)& 6\\ & \text{are drawn to the circle}{x}^2+y^2-6\sqrt{3}x-6y+27& & \\ & \text{= 0 from the origin, then}2\sqrt{3}\tan \theta \text{is equal to}& & \\ \left(c\right)& \text{The shortest distance between parabolas}{y}^2=4x& \left(r\right)& 2\sqrt{7}\\ & \text{and}{y}^2=2x-6\text{is d, then}{d}^2=& & \\ \left(d\right)& \text{Distance between foci of the curve represented}& \left(s\right)& 1\\ & \text{by the equation}x=1+4\cos \theta ,\text{and}y=2+3\sin \theta \text{is}& & \end{array}$

• A. (a) – (r); (b) – (s); (c) – (p); (d) – (q)
• B. (a) – (s); (b) – (r); (c) – (p); (d) – (r)
• C. (a) – (s); (b) – (q); (c) – (p); (d) – (r)
• D. (a) – (r); (b) – (p); (c) – (q); (d) – (s)

Answer 1

The correct option is C (a) – (s); (b) – (q); (c) – (p); (d) – (r)

(A) Slope of line joining $(t-1,2t+2)$ and $(2t+1,t)$ is:

$\frac{2t+2-t}{t-1-2t-1}=\frac{t+2}{-(t+2)}=-1$

Original line is perpendicular to the line joining these two points.

So, slope of original line is $1$


(B)Let $y=mx$ be the tangent drawn from origin. Then

$3=\frac{\left|\begin{array}{c}3-3\sqrt{3}m\end{array}\right|}{\sqrt{1+m^2}}$

$\Rightarrow 1+m^2=(1-\sqrt{3}m{)}^2$

$\Rightarrow 1+m^2=1+3m^2-2\sqrt{3}m$

$\Rightarrow 2m^2-2\sqrt{3}m=0\Rightarrow m=0,\sqrt{3}$

So, to find angle between tangents,

$\tan \theta =\left|\begin{array}{c}\frac{m_1-m_2}{1+m_1{m}_2}\end{array}\right|=\sqrt{3}$

Hence, $2\sqrt{3}\tan \theta =6$


(C)$y^2=4x$ ... (1)

$y^2=2(x-3)$ ... (2)

Shortest distance lies along the common normal

Equation of normal to (1) is

$y=mx-2m-m^3$ ... (3)
Point of contact is $(m^2,-2m)$

Equation of normal to (2) is

$y=m(x-3)-m-\frac{m^3}{2}$ ... (4)
Point of contact is $(3+\frac{m^2}{2},-m)$

if (3) & (4) represent same line then,

$-2m-m^3=-3m-m-\frac{m^3}{2}$

$\Rightarrow 2m-\frac{m^3}{2}=0$

$\Rightarrow m=0,\pm 2$

So, foot of normals are

(1) $(0,0)\&(3,0)$

(2) $(4,-4)\&(5,-2)$

(3) $(4,4)\&(5,2)$

So, possible values of $d=3,\sqrt{5}$

So $d_{min}=\sqrt{5}\Rightarrow d^2=5$


(D) Equation of curve is $\frac{(x-1{)}^2}{16}+\frac{(y-2{)}^2}{9}=1$

Distance between the focii is

$=2ae$ where $e$ is the eccentricity which is equal to $\sqrt{1-\frac{b^2}{a^2}}$

$\Rightarrow 2ae=2\sqrt{a^2-b^2}=2\sqrt{16-9}$

$=2\sqrt{7}$