Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q.
Column I Column II
A The minimum and maximum distance of a point $(2,6)$ from the ellipse are $9 x^2+8 y^2-36 x-16 y-28=0$ p 0
B The minimum and maximum distance of a point $\left(\frac{9}{5}, \frac{12}{5}\right)$ from the ellipse $4(3 x+4 y)^2+9(4 x-3 y)^2=900$ are q 2
C If $E: 2 x^2+y^2=2$ 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 $r _1, r _2, r _3 \ldots$ are the radii of $C _1, C _2, C _3 \ldots$ 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 $x^2+4 y^2=16$ with coordinate axes is s 8

Conic Sections

Solution:

(A)$9\left(x^2-4 x+4\right)+8\left(y^2-2 y+1\right)=28+36+8$
$\Rightarrow 9(x-2)^2+8(y-1)^2=72$
$\Rightarrow \frac{(x-2)^2}{8}+\frac{(y-1)^2}{9}=1$
image
Minimum distance of $(2,6)$ from the ellipse is $2 \&$ maximum distance of $(2,6)$ from the ellipse is 8
(B) $\frac{(3 x+4 y)^2}{225}+\frac{(4 x-3 y)^2}{100}=1$
$\frac{\left(\frac{3 x+4 y}{5}\right)^2}{9}+\frac{\left(\frac{4 x-3 y}{5}\right)^2}{4}=1$
image
Point $\left(\frac{9}{5}, \frac{12}{5}\right)$ lie on line $4 x-3 y=0$
$\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 $(3.6 .12)^{1 / 3}=6$
(D) $\frac{x^2}{16}+\frac{y^2}{4}=1$
image
equation of tangent at $P$ is
$\frac{x \cos \theta}{4}+\frac{y \sin \theta}{2}=1$
$x \cos \theta+2 y \sin \theta=4$
Area of triangle
$\Delta=\frac{1}{2} \cdot 4 \sec \theta \cdot 2 \text{cosec} \theta=\frac{8}{\sin 2 \theta} $
minimum area $=8 \text { when } \sin 2 \theta=1$