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

The correct option is C $\frac{(x+2{)}^2}{40}+\frac{y^2}{49}=1$

$A(-2,-3),B(-2,3)$
Let the third vertex be C(x,y)
AB + BC+CA=20
BC+CA=14
So locus of C is ellipse with A,B as focii.
2a=14
2ae=6
$e=\frac{3}{7}$
$\therefore$ we get b as $\sqrt{40}$
$\therefore$ Equation of ellipse is $\frac{(x+2{)}^2}{40}+\frac{y^2}{49}=1$