Question 67

Mathematics Hyperbola Hard

The eccentric angle of the extremities of latus-rectum of the ellipse \(\frac{{x}^2}{{a}^2}^{}+\frac{{y}^2}{{b}^2}^{}=1\) are given by

(A) \( tan ^{-1}(\pm\frac{ae}{b})\)
(B) \( tan ^{-1}(\pm\frac{be}{e})\)
(C) \( tan ^{-1}(\pm\frac{b}{ae})\)
(D) \( tan ^{-1}(\pm\frac{a}{be})\)
✓ Correct Answer: Option C Verified Solution

📘 Step-by-Step Solution & Detailed Explanation:

We know that the length of the latus rectum is: \[ \frac{2b^2}{a} \] and the latus rectum is perpendicular to the major axis and passes through the focus. Hence, the coordinates of one of its ends are: \[ \left( ae, \frac{b^2}{a} \right) \] If the eccentric angle is \( \theta \), then the coordinates of one end of the latus rectum are: \[ (a \cos \theta, b \sin \theta) \] Hence, we have: \[ a \cos \theta = ae, \quad b \sin \theta = \frac{b^2}{a} \] From this, we get: \[ \tan \theta = \frac{b}{ae} \] Since there are two ends of the latus rectum, the eccentric angle \( \theta \) is: \[ \theta = \tan^{-1}\left( \pm \frac{b}{ae} \right) \]

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