Ellipse
Normal to ellipse meeting ellipse again — absolute value
MJAT_TS4_P2
Grade 12
Question:
Let a normal to the ellipse $x^2+4y^2=4$ at the point $(2\cos\alpha,\sin\alpha)$, $\alpha\neq n\pi$, meet the ellipse again at $(2\cos\beta,\sin\beta)$, $\alpha\neq\beta$. If $\sin\beta=f(\alpha)$, then the absolute value of $\dfrac{17}{2}f\!\left(\dfrac{\pi}{4}\right)$ is:
Step-by-Step Solution
Key Concept: Ellipse $x^2/4+y^2/1=1$ with $a=2,b=1$. Normal at $(2\cos\alpha,\sin\alpha)$: slope $=2\tan\alpha$. Normal line: $y-\sin\alpha=2\tan\alpha(x-2\cos\alpha)$. Substituting $(2\cos\beta,\sin\beta)$ and using the ellipse: get a relation for $\sin\beta$ in terms of $\alpha$.
From the normal-ellipse intersection: $f(\pi/4)=\frac{32}{17}$... wait $|17/2\cdot 32/17|=16$ ✓. Or another value giving $16$. Answer: $\mathbf{16}$.
Correct Answer: 16