The locus of mid-points of the line segments joining - 3 , - 5 and the points on the ellipse x 2 4 + y 2 9 = 1 is :

Question

Aicharya · Accepted Answer

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

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

Then, by using mid-point formula, we get

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

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

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

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

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

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

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

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

The locus of mid-points of the line segments joining - 3 , - 5 and the points on the ellipse x 2 4 + y 2 9 =…