What is the area of the polygon whose vertices are the points of intersection of the curves $x^2 + y^2 =25$ and $(x-4)^2 + 9y^2 = 81$? $\textbf{(A)}\ 24\qquad$ $\textbf{(B)}\ 27\qquad$ $\textbf{(C)}\ 36\qquad$ $\textbf{(D)}\ 37.5\qquad$ $\textbf{(E)}\ 42$
Solution
From $x^2+y^2=25$: $y^2=25-x^2$. Substituting into $(x-4)^2+9y^2=81$ gives $(x-4)^2+9(25-x^2)=81$, i.e. $-8x^2-8x+241=81$, so $x^2+x-20=0$ and $(x+5)(x-4)=0$.
$x=-5$ gives $y=0$; $x=4$ gives $y=\pm3$. The three intersection points are $(-5,0),(4,3),(4,-3)$.
This triangle has base $6$ (from $(4,3)$ to $(4,-3)$) and height $9$ (the horizontal distance from $(-5,0)$ to the line $x=4$).
