Question

Let $A,B$ be one of the ends of major and minor axes of an ellipse $\frac{x^2}{16}+\frac{y^2}{9}=1$ respectively, where $A$ lies on positive $x-$ axis and $B$ lies on positive $y-$ axis. Let $L_1$ be a vertical line through $A$ and $L_2$ be a horizontal line through $B$, also $L_1$ and $L_2$ intersect at the point $P.$ If $S=0$ is a circle having $OP$ as radius, where $O$ is the origin and $P$ is the centre, then which of the following is/are correct?

• A. $S=0$ intersects $x-$axis at $(8,0)$
• B. $S=0$ intersects $y-$axis at $(0,\pm 5)$
• C. $S=0$ intersects the ellipse at two points only.
• D. $S=0$ does not intersects the ellipse.

Answer 1

Answer: (A), (C)

The correct options are
A $S=0$ intersects $x-$axis at $(8,0)$
C $S=0$ intersects the ellipse at two points only.


Given ellipse: $\frac{x^2}{16}+\frac{y^2}{9}=1$
Here,
$a^2=16,b^2=9A\equiv (4,0),B\equiv (0,3)$
Now, the equation of $L_1$ is $x=4$
Equation of $L_2$ is $y=3$
$P\equiv (4,3)$
$OP=\sqrt{4^2+3^2}=5$

Equation of circle having $OP$ as radius and Centre at $P$ is
$(x-4{)}^2+(y-3{)}^2=5^2$
$\Rightarrow S\equiv x^2+y^2-8x-6y=0$

$y=0\Rightarrow x^2-8x=0\Rightarrow x=0,8x=0\Rightarrow y^2-6y=0\Rightarrow y=0,6$
The circle and ellipse intersect at $2$ points only.