Mathematics - Ellipse Question with Solution | TestHub

Let the two given vertices be $A(-2,-3)$ and $B(-2,3)$, and the third vertex be $P(x,y)$.

The distance $AB=\sqrt{(-2-(-2){)}^2+(3-(-3){)}^2}=\sqrt{0^2+6^2}=6$.

The perimeter is $PA+PB+AB=20$.

So, $PA+PB=20-AB=20-6=14$.

This is the definition of an ellipse where $A$ and $B$ are the foci, and $2a=14$, so $a=7$.

The distance between foci is $2c=AB=6$, so $c=3$.

For an ellipse, $b^2=a^2-c^2=7^2-3^2=49-9=40$.

The center of the ellipse is the midpoint of $AB$, which is $\left(\frac{-2+(-2)}{2},\frac{-3+3}{2}\right)=(-2,0)$.

Since the foci lie on a vertical line ($x=-2$), the major axis is vertical.

The equation of the ellipse is $\frac{(x-h{)}^2}{b^2}+\frac{(y-k{)}^2}{a^2}=1$.

Substituting the values, we get $\frac{(x-(-2){)}^2}{40}+\frac{(y-0{)}^2}{49}=1$.

Thus, the locus of the third vertex is $\frac{(x+2{)}^2}{40}+\frac{y^2}{49}=1$.