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)

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

Let the coordinates of the third vertex be $A:(x,y)$

Given that perimeter is $20$

The other two vertices given are $B:(-2,-3)$ and $C:(-2,3)$

We know that Perimeter $=AB+AC+BC$ ........(1)

Using distance formula, we get

$\Rightarrow AB=\sqrt{{(x+2)}^2+{(y+3)}^2}$

$\Rightarrow AC=\sqrt{{(x+2)}^2+{(y-3)}^2}$

$\Rightarrow BC=\sqrt{36}=6$

Putting these values in (1), we get

$\sqrt{{(x+2)}^2+{(y+3)}^2}+\sqrt{{(x+2)}^2+{(y-3)}^2}=20-6=14$

$\Rightarrow \sqrt{{(x+2)}^2+{(y+3)}^2}=14-\sqrt{{(x+2)}^2+{(y-3)}^2}$

Squaring both sides, we get

${\Rightarrow (x+2)}^2+{(y+3)}^2=196+{(x+2)}^2+{(y-3)}^2-28\sqrt{{(x+2)}^2+{(y-3)}^2}$

$\Rightarrow 12y-196=-28\sqrt{{(x+2)}^2+{(y-3)}^2}$

$\Rightarrow 49-3y=7\sqrt{{(x+2)}^2+{(y-3)}^2}$

Again squaring both sides, we get

$\Rightarrow 2401+9y^2-294y=49{(x+2)}^2+49y^2+441-294y$

$\Rightarrow 1960=49{(x+2)}^2+40y^2$

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