## Calculus 10th Edition

Two horizontal tangents at $(-4, 0)$ and at $(-4, 10)$. Two vertical tangents at $(0, 5)$ and at $(-8, 5.)$
$\frac{d}{dx}(25x^2)+\frac{d}{dx}(16y^2)+\frac{d}{dx}(200x)+\frac{d}{dx}(-160y)+\frac{d}{dx}(400)=0$ $50x+\dfrac{dy}{dx}(32y)+200-\dfrac{dy}{dx}(160)=0$ $\dfrac{dy}{dx}=\dfrac{25x+100}{80-16y}$. Case I: Horizontal tangent $\rightarrow$ $\dfrac{dy}{dx}=0\rightarrow25x+100=0\rightarrow x=-4$ Substituting $x=-4$ into the original equation gives $25(-4)^2+16y^2+200(-4)+(-160y)+400$ $=16y(y-10)=0\rightarrow y=0$ or $y=10\rightarrow$ horizontal tangent at $(-4, 0)$ and at $(-4, 10)$. Case II: Vertical tangent $\rightarrow \dfrac{dy}{dx}=$undefined$\rightarrow$ $80-16y=0\rightarrow y=5$ Substituting $y=5$ into the original equation gives: $25x^2+16(5)^2+200x-160(5)+400=$ $25x(x+8)=0\rightarrow x=0$ or $x=-8\rightarrow$ vertical tangent at $(0, 5)$ and $(-8, 5)$.