Question

If the angle between the straight lines joining the foci and the ends of the minor axis of the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \] is \(90^\circ\), then its eccentricity is

• A. \(\frac{1}{2}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{1}{3}\)
• D. \(\frac{1}{\sqrt{2}}\)

Hint:

For an ellipse, always use \(b^2=a^2(1-e^2)\). If two lines are perpendicular, use the dot product condition \(\vec{A}\cdot\vec{B}=0\).

Answer 1

The Correct Option is D

Step 1: Identify key points.
For $x^2/a^2+y^2/b^2=1$ ($a>b>0$): foci at $(\pm ae,0)$ and ends of minor axis at $(0,\pm b)$.
Step 2: Form direction vectors from minor axis end to foci.
From $B=(0,b)$: vectors $\vec{u}=(ae,-b)$ to $F_1=(ae,0)$ and $\vec{v}=(-ae,-b)$ to $F_2=(-ae,0)$.
Step 3: Apply perpendicularity.
$\vec{u}\cdot\vec{v}=0$: $(ae)(-ae)+(-b)(-b)=0 \implies b^2=a^2e^2$.
Step 4: Use $b^2=a^2(1-e^2)$.
$a^2(1-e^2)=a^2e^2 \implies 1=2e^2 \implies e=1/\sqrt{2}$.
Step 5: Verify.
$e=1/\sqrt{2}\approx 0.707<1$, valid for an ellipse.
Step 6: State the answer.
\[ \boxed{e=\frac{1}{\sqrt{2}}} \]