Question

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is

• A. $36x^2+16y^2+90x+56y+145=0$
• B. $9x^2+4y^2+18x+8y+145=0$
• C. $36x^2+16y^2+72x+32y+145=0$
• D. $36x^2+16y^2+108x+80y+145=0$

Answer 1

The correct option is D $36x^2+16y^2+108x+80y+145=0$

Let general point on ellipse $(2\cos \theta ,3\sin \theta )$ be
Let mid-point of $(2\cos \theta ,3\sin \theta )$ and $(-3,-5)$ be $(h,k)$
$\therefore 2\cos \theta -3=2h$ and $3\sin \theta -5=2k$
$\Rightarrow \cos \theta =\frac{2h+3}{2}$ and $\sin \theta =\frac{2k+5}{3}$
We know that ${\sin }^2\theta +{\cos }^2\theta =1$
Putting the value of $\sin \theta$ and $\cos \theta$ by replacing $h$ with $x$ and $k$ with $y$
$\Rightarrow {\left(\frac{2y+5}{3}\right)}^2+{\left(\frac{2x+3}{2}\right)}^2=1$
$\Rightarrow 4(2y+5{)}^2+9(2x+3{)}^2=36$
$\Rightarrow 36x^2+16y^2+108x+80y+145=0$