Analytic Geometry – Problem 22: Consider the ellipse with left/right foci and left/right vertices

Consider the ellipse $C:\frac{x^2}{25}+\frac{y^2}{9}=1$ with left/right foci $F_1,F_2$ and left/right vertices $A,B$. Point $P$ moves on the ellipse. Which statement is false?

A. The eccentricity is $e=\frac45$.

B. The perimeter of $\triangle F_1PF_2$ is 18.

C. The product of the slopes of lines $PA$ and $PB$ is the constant $-\frac{9}{25}$.

D. If $\angle F_1PF_2=90^\circ$, then the area of $\triangle F_1PF_2$ is 8.

For $\frac{x^2}{25}+\frac{y^2}{9}=1$, we have $a=5$, $b=3$, so $c^2=a^2-b^2=16\Rightarrow c=4$.

A. $e=\frac{c}{a}=\frac45$ is true.

B. For any $P\in C$, $|PF_1|+|PF_2|=2a=10$ and $|F_1F_2|=2c=8$. Hence $$\text{perimeter}(\triangle F_1PF_2)=|PF_1|+|PF_2|+|F_1F_2|=10+8=18,$$ so B is true.

C. Write $A(-5,0)$, $B(5,0)$, $P(x,y)$. Then $$\text{slope}(PA)=\frac{y}{x+5},\qquad \text{slope}(PB)=\frac{y}{x-5}.$$ Their product is $$\frac{y^2}{x^2-25}.$$ From the ellipse equation, $y^2=9\left(1-\frac{x^2}{25}\right)=\frac{9(25-x^2)}{25}$, so $$\frac{y^2}{x^2-25}=\frac{\frac{9(25-x^2)}{25}}{x^2-25}=-\frac{9}{25}.$$ Thus C is true.

D. If $\angle F_1PF_2=90^\circ$, then triangle $F_1PF_2$ is right at $P$, so $$PF_1^2+PF_2^2=F_1F_2^2=8^2=64.$$ Also $PF_1+PF_2=10$. Hence $$(PF_1+PF_2)^2=PF_1^2+PF_2^2+2PF_1PF_2\Rightarrow 100=64+2PF_1PF_2,$$ so $PF_1PF_2=18$. The area is $$S_{\triangle F_1PF_2}=\frac12\,PF_1\cdot PF_2=\frac12\cdot 18=9\ne 8.$$ Therefore statement D is false.

The false statement is D.