The locus of mid-points of the line segments joining (- 3, - 5) and...

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

Let, the mid-point of line segments joining \left(- 3, - 5\right)$and \left(2sin\theta, 3cos\theta\right)$ is \left(h, k\right)$</p><p>Then, by using mid-point formula, we get</p><p>$\dfrac{2sin\theta - 3}{2} = h, \dfrac{3cos\theta - 5}{2} = k$</p><p>$⇒ 2sin\theta = 2h + 3, 3cos\theta = 2k + 5$</p><p>$⇒ sin\theta = \dfrac{2h + 3}{2}, cos\theta = \dfrac{2k + 5}{3}$</p><p>We know,$\sin^{2}\theta + \cos^{2}\theta = 1$</p><p>$⇒ {\left(\dfrac{2h + 3}{2}\right)}^{2} + {\left(\dfrac{2k + 5}{3}\right)}^{2} = 1$</p><p>$⇒ \dfrac{1}{4}\left[4{h}^{2} + 9 + 12h\right] + \dfrac{1}{9}\left[4{k}^{2} + 25 + 20k\right] = 1$</p><p>$⇒ 36{h}^{2} + 16{k}^{2} + 108h + 80k + 145 = 0$</p><p>Hence, the locus of \left(x, y\right)$ is $36{x}^{2} + 16{y}^{2} + 108x + 80y + 145 = 0$