Question

The perimeter of a triangle is 20 and the points (-2, -3) and (-2, 3) are two of the vertices of it. Then the locus of third vertex is:

• A. $\frac{(x - 2)^{2}}{49} + \frac{y^{2}}{40} = 1$
• B. $\frac{(x + 2)^{2}}{49} + \frac{y^{2}}{40} = 1$
• C. $\frac{(x + 2)^{2}}{40} + \frac{y^{2}}{49} = 1$
• D. $\frac{(x - 2)^{2}}{40} + \frac{y^{2}}{49} = 1$

Answer 1

Answer: (C)

$\frac{(x + 2)^{2}}{40} + \frac{y^{2}}{49} = 1$

Answer 2

Let the third vertex be

$\sqrt{(h + 2)^{2} + (k + 3)^{2}} + \sqrt{(h + 2)^{2} + (k - 3)^{2}} + 6 = 20$.

Squaring and simplifying, we get 49 (h + 2)² + 40k² = 40 x 49.

⇒ $\frac{(h + 2)^{2}}{40} + \frac{k^{2}}{49} = 1$

So, locus of (h, k) is

⇒ $\frac{(x + 2)^{2}}{40} + \frac{y^{2}}{49} = 1$.