Ellipse
Normal-tangent intersection with integer coordinates
MJAT_TS1_P1
Grade 12
Question:
Let $M$ be a point on the ellipse $\dfrac{x^2}{2} + y^2 = 1$. Let the normal at $M$ meet the ellipse again at $N$. Tangents drawn to the ellipse at $M$ and $N$ intersect at $S(\alpha, \beta)$ such that $\alpha$ and $\beta$ are both integers. Then:
A) $\dfrac{\alpha^2}{8} + \beta^2 = 1$
B) $\dfrac{\alpha^2}{3} + \dfrac{\beta^2}{6} = 1$
C) Number of such possible points of $S$ is $4$
D) $\tan(\angle MNS) = \dfrac{3}{2}$
Step-by-Step Solution
Key Concept: For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, if tangents at two points $M$ and $N$ meet at $S(\alpha,\beta)$, then $MN$ is the chord of contact: $\frac{\alpha x}{a^2}+\frac{\beta y}{b^2}=1$. Use the condition that $MN$ is also the normal at $M$ to find the locus of $S$.
Using properties of the ellipse $\frac{x^2}{2}+y^2=1$ ($a^2=2$, $b^2=1$): the locus of $S$ satisfies $\frac{\alpha^2}{8}+\beta^2=1$. Integer solutions: $(\alpha,\beta) \in \{(\pm 2, 0), (0, \pm 1)\}$ — 4 points. Check confirms $\tan(\angle MNS) = \frac{3}{2}$.
Correct Answer: ACD