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 the foci and ends of minor axis.
For the ellipse \[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \] the foci are \[ (ae,0)\quad \text{and}\quad (-ae,0) \] The ends of the minor axis are \[ (0,b)\quad \text{and}\quad (0,-b) \]

Step 2: Use one end of the minor axis.
Consider the end of the minor axis \[ (0,b) \] The vectors from \((0,b)\) to the foci are \[ \vec{A}=(ae,-b) \] and \[ \vec{B}=(-ae,-b) \]

Step 3: Apply perpendicular condition.
Since the angle between the two lines is \(90^\circ\), their dot product must be zero.
So, \[ \vec{A}\cdot \vec{B}=0 \] \[ (ae)(-ae)+(-b)(-b)=0 \] \[ -a^2e^2+b^2=0 \] Hence, \[ b^2=a^2e^2 \]

Step 4: Use the ellipse eccentricity relation.
For an ellipse, \[ b^2=a^2(1-e^2) \] Using \[ b^2=a^2e^2, \] we get \[ a^2e^2=a^2(1-e^2) \] Dividing by \(a^2\), \[ e^2=1-e^2 \] \[ 2e^2=1 \] \[ e^2=\frac{1}{2} \] Therefore, \[ e=\frac{1}{\sqrt{2}} \]

Step 5: Final conclusion.
Therefore, the eccentricity of the ellipse is \[ \boxed{\frac{1}{\sqrt{2}}} \]