Mathematics - Ellipse Question with Solution | TestHub

Parametric point on the given ellipse is $\left(2\sin \theta , 3\cos \theta \right)$

Let, the mid-point of line segments joining $\left(-3,-5\right)$ and $\left(2\sin \theta ,3\cos \theta \right)$ is $\left(h, k\right)$

Then, by using mid-point formula, we get

$\frac{2\sin \theta -3}{2}=h, \frac{3\cos \theta -5}{2}=k$

$\Rightarrow 2\sin \theta =2h+3, 3\cos \theta =2k+5$

$\Rightarrow \sin \theta =\frac{2h+3}{2}, \cos \theta =\frac{2k+5}{3}$

We know, ${\sin }^2\theta +{\cos }^2\theta =1$

$\Rightarrow {\left(\frac{2h+3}{2}\right)}^2+{\left(\frac{2k+5}{3}\right)}^2=1$

$\Rightarrow \frac{1}{4}\left[4h^2+9+12h\right]+\frac{1}{9}\left[4k^2+25+20k\right]=1$

$\Rightarrow 36h^2+16k^2+108h+80k+145=0$

Hence, the locus of $\left(x,y\right)$ is $36x^2+16y^2+108x+80y+145=0$