Question

Column I		Column II
A	The minimum and maximum distance of a point $(2,6)$ from the ellipse are $9x^2+8y^2-36x-16y-28=0$	p	0
B	The minimum and maximum distance of a point $\left(\frac{9}{5},\frac{12}{5}\right)$ from the ellipse $4(3x+4y{)}^2+9(4x-3y{)}^2=900$ are	q	2
C	If $E:2x^2+y^2=2$ and director circle of $E$ is $C_1$, director circle of $C_1$ is $C_2$ director circle of $C_2$ is $C_3$ and so on. If $r_1,r_2,r_3\dots$ are the radii of $C_1,C_2,C_3\dots$ respectively then G.M. of $r_1^2,r_2^2,r_3^2$ is	r	6
D	Minimum area of the triangle formed by any tangent to the ellipse $x^2+4y^2=16$ with coordinate axes is	s	8

• A. $(A)\to (q,s);(B)\to (p,r);(C)\to (r);(D)\to (s)$
• B. $(A)\to (s,s);(B)\to (p,q);(C)\to (r);(D)\to (s)$
• C. $(A)\to (q,p);(B)\to (r,r);(C)\to (r);(D)\to (s)$
• D. $(A)\to (p,s);(B)\to (p,q);(C)\to (r);(D)\to (s)$

Answer 1

Answer: (A)

(A)$9\left(x^2-4x+4\right)+8\left(y^2-2y+1\right)=28+36+8$
$\Rightarrow 9(x-2{)}^2+8(y-1{)}^2=72$
$\Rightarrow \frac{(x-2{)}^2}{8}+\frac{(y-1{)}^2}{9}=1$

Minimum distance of $(2,6)$ from the ellipse is $2\&$ maximum distance of $(2,6)$ from the ellipse is 8
(B) $\frac{(3x+4y{)}^2}{225}+\frac{(4x-3y{)}^2}{100}=1$
$\frac{{\left(\frac{3x+4y}{5}\right)}^2}{9}+\frac{{\left(\frac{4x-3y}{5}\right)}^2}{4}=1$

Point $\left(\frac{9}{5},\frac{12}{5}\right)$ lie on line $4x-3y=0$
$\therefore$ Minimum distance$=0$
Maximum distance $=6$
(C)$C_1:x^2+y^2=3$
$C_2:x^2+y^2=6$
$C_3:x^2+y^2=12$
$GM$ of $3,6,12$ is $(\mathrm{3.6.12}{)}^{1/3}=6$
(D) $\frac{x^2}{16}+\frac{y^2}{4}=1$

equation of tangent at $P$ is
$\frac{x\cos \theta }{4}+\frac{y\sin \theta }{2}=1$
$x\cos \theta +2y\sin \theta =4$
Area of triangle
$\Delta =\frac{1}{2}\cdot 4\sec \theta \cdot 2\text{cosec}\theta =\frac{8}{\sin 2\theta }$
minimum area $=8\text{ when }\sin 2\theta =1$