Question
If the tangent to the ellipse $x^{2}+4 y^{2}=16$ at the point with eccentric angle $\phi$ is a normal to the curve $x^{2}+y^{2}-8 x-4 y$ $=0$, then $\phi$ equals(a) $\frac{\pi}{3}$(b) $\frac{-\pi}{3}$(c) $\frac{\pi}{6}$(d) $\frac{\pi}{2}$
Step 1
We can rewrite this in standard form by dividing both sides by 16, which gives us $\frac{x^{2}}{16} + \frac{y^{2}}{4} = 1$. This is in the form $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$, where $a^{2} = 16$ and $b^{2} = 4$. Show more…
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