Ellipse & Parabola
Eccentricity from normal-tangent parallelism condition
MJAT_TS2_P1
Grade 12
Question:
Let $a$, $b$, $k$ be positive real numbers. Suppose $Q$ is a point on the parabola $y^2=4kx$ such that its distance from the focus is $2k$ and it lies in the first quadrant. An ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ passes through $Q$. If the normal to the parabola at $Q$ and the tangent to the ellipse at $Q$ are parallel, then the eccentricity of the ellipse is:
A) $\dfrac{\sqrt{2}}{3}$
B) $\dfrac{2}{\sqrt{3}}$
C) $\dfrac{1}{\sqrt{2}}$
D) $\dfrac{1}{\sqrt{3}}$
Step-by-Step Solution
Key Concept: Focus of $y^2=4kx$ is $(k,0)$. Distance from $Q$ to focus $=2k$ means $x_Q+k=2k\Rightarrow x_Q=k$. On parabola: $y_Q^2=4k^2\Rightarrow y_Q=2k$ (first quadrant). So $Q=(k,2k)$. Slope of normal at $Q$: $m_N=-1$ (since $dy/dx|_Q = k/y_Q = 1/2$, so $m_N=-1$).
$b^2/(2a^2)=1\Rightarrow b^2=2a^2$. Since $b>a$, the ellipse is vertical (major axis along $y$). For vertical ellipse: $a^2=b^2(1-e^2)\Rightarrow a^2=2a^2(1-e^2)\Rightarrow e^2=1/2\Rightarrow e=1/\sqrt{2}$.
Correct Answer: C