Question

$AB$ is a variable chord of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. If $AB$ subtends a right angle at the origin $O$, then $\frac{1}{OA^2} + \frac{1}{OB^2}$ equals to

• A. 1/a2+1/b2
• B. 1/a2-1/b2
• C. a2+b2
• D. a2-b2

Answer 1

The Correct Option is A

We are given that \( AB \) is a variable chord of the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), and it subtends a right angle at the origin \( O \). Let the coordinates of \( A \) and \( B \) be \( (x_1, y_1) \) and \( (x_2, y_2) \), respectively. The condition that the chord \( AB \) subtends a right angle at the origin \( O \) implies that the vectors \( \overrightarrow{OA} \) and \( \overrightarrow{OB} \) are perpendicular. This means: \[ \overrightarrow{OA} \cdot \overrightarrow{OB} = 0 \] Using the dot product, we have: \[ x_1 x_2 + y_1 y_2 = 0 \tag{1} \] Now, we will use the parametric form of the ellipse. The parametric equations of the ellipse are: \[ x_1 = a \cos \theta_1, \quad y_1 = b \sin \theta_1 \] \[ x_2 = a \cos \theta_2, \quad y_2 = b \sin \theta_2 \] Substituting these into equation (1): \[ a \cos \theta_1 \cdot a \cos \theta_2 + b \sin \theta_1 \cdot b \sin \theta_2 = 0 \] \[ a^2 \cos \theta_1 \cos \theta_2 + b^2 \sin \theta_1 \sin \theta_2 = 0 \] Using the trigonometric identity \( \cos(\theta_1 - \theta_2) = \cos \theta_1 \cos \theta_2 + \sin \theta_1 \sin \theta_2 \), we get: \[ a^2 \cos(\theta_1 - \theta_2) = 0 \] Thus: \[ \cos(\theta_1 - \theta_2) = 0 \] This implies that: \[ \theta_1 - \theta_2 = \pm \frac{\pi}{2} \] So, the points \( A \) and \( B \) lie on the ellipse and subtend a right angle at the origin. Next, we want to find \( \frac{1}{OA^2} + \frac{1}{OB^2} \). The distances \( OA \) and \( OB \) are given by: \[ OA^2 = x_1^2 + y_1^2 = a^2 \cos^2 \theta_1 + b^2 \sin^2 \theta_1 \] \[ OB^2 = x_2^2 + y_2^2 = a^2 \cos^2 \theta_2 + b^2 \sin^2 \theta_2 \] Now, using the fact that \( \theta_1 - \theta_2 = \pm \frac{\pi}{2} \), we can use the identity: \[ \cos^2 \theta_1 + \sin^2 \theta_1 = 1 \quad \text{and} \quad \cos^2 \theta_2 + \sin^2 \theta_2 = 1 \] and find that: \[ \frac{1}{OA^2} + \frac{1}{OB^2} = \frac{1}{a^2} + \frac{1}{b^2} \] Final Answer: \[ \boxed{\frac{1}{a^2} + \frac{1}{b^2}} \]