Question

An ellipse whose major axis is parallel to $x-$axis is such that the segments of a focal chord are of $1$ and $3$ units. The lines $ax+by+c=0$ are the chords of the ellipse such that $a,b,c$ are in A.P. and bisected by the point at which they are concurrent. The equation of auxiliary circle is $x^2+y^2+2\alpha x+2\beta y-2\alpha -1=0.$ Then

• A. The locus of perpendicular tangents to the ellipse is $(x-1{)}^2+(y+2{)}^2=7$
• B. The locus of perpendicular tangents to the ellipse is $(x-1{)}^2+(y+2{)}^2=19$
• C. Area of auxiliary circle is $2\pi$ sq. unit
• D. Eccentricity of the ellipse is $\frac{1}{2}$

Answer 1

Answer: (A), (D)

Eccentricity of the ellipse is $\frac{1}{2}$

$a,b,c$ are in A.P.
$\Rightarrow a-2b+c=0$
So family of chord lines $ax+by+c=0$ are concurrent at $(1,-2)$
$\therefore$ Centre of auxiliary circle $x^2+y^2+2\alpha x+2\beta y-2\alpha -1=0$ is $(-\alpha ,-\beta )=(1,-2)$
So, radius of auxiliary circle $=2$ unit
Area $=\pi \times 2^2=4\pi$ sq. unit
$\Rightarrow$ Length of major axis $=2a=4$ unit

Using focal chord property,$\because \frac{1}{SP}+\frac{1}{SQ}=\frac{2a}{b^2}$
$\Rightarrow b=\sqrt{3}$
Hence, $e=\frac{1}{2}$ and equation of director circle is
$(x-1{)}^2+(y+2{)}^2=4+3\Rightarrow (x-1{)}^2+(y+2{)}^2=7$