Question
If the normal to the ellipse $\frac{x^{2}}{14}+\frac{y^{2}}{5}=1$ at the point $\theta$ intersects the curve again at the point $2 \theta$, prove that $\sec \theta=-\frac{3}{2}$.
Step 1
We can write this in parametric form as $x = \sqrt{14}\cos\theta$ and $y = \sqrt{5}\sin\theta$. Show more…
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PARAMETRIC EQUATIONS AND POLAR COORDINATES
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