Question

The tangent at point(a cosθ, b sinθ), 0<θ<\(\frac{\pi}{2}\), to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) meets the x-axis at T and y-axis at T₁, Then the value of \(\min_{\,\,\,0<\theta<\frac{\pi}{2}}\)(OT)(OT₁) is

• A. ab
• B. 2ab
• C. 0
• D. 1

Answer 1

The Correct Option is B

Tangent Equation to the Ellipse (I): The given equation represents the equation of a tangent line to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) at the point (acosθ, bsinθ) on the ellipse.

Equation (I): \(x \times a\cosθ + y \times b\sinθ = 1\)

Joint Equation of Lines Joining Points to the Origin: The equation describes the joint equation of lines that connect the points of intersection of the tangent (I) with the auxiliary circle \(x^2 + y^2 = a^2\) to the origin, which is the center of the circle.

The equation is: \(x^2 + y^2 = a^2 \times [x \times a\cosθ + y \times b\sinθ]^2\)

Condition for Lines at Right Angles: The next step involves finding the condition for these lines to be at right angles to each other. This condition is achieved when the coefficients of \(x^2\) and \(y^2\) in the equation are such that their sum is zero.

The derived equation for this condition is: \(1 - a^2 \times (\frac{\cos^2θ}{a^2}) + 1 - a^2 \times (\frac{\sin^2θ}{b^2}) = 0\)

Solving for Eccentricity (e): The equation for the condition of right angles is then simplified and rearranged to solve for the eccentricity (e) of the ellipse.

The final equation is: \(e = (1 + \sin^2θ)^{(\frac{-1}{2})}\)

The correct answer is option (B) : 2ab