Ellipse
Tangents to ellipse
Grade 11
Question:
<p>Tangents are drawn to the ellipse \(\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1\) \((a > b)\) and the circle \(x^2 + y^2 = a^2\) at the points where a common ordinate cuts them (on the same side of the \(x\)-axis). Then the greatest acute angle between these tangents is given by</p>
<p>\(\tan^{-1}\left(\dfrac{a-b}{2\sqrt{ab}}\right)\)</p>
<p>\(\tan^{-1}\left(\dfrac{a+b}{2\sqrt{ab}}\right)\)</p>
<p>\(\tan^{-1}\left(\dfrac{2ab}{\sqrt{a-b}}\right)\)</p>
<p>\(\tan^{-1}\left(\dfrac{2ab}{\sqrt{a+b}}\right)\)</p>
Step-by-Step Solution
Key Concept: For a common ordinate (vertical line x = c), find tangent slopes at the ellipse and circle, then maximize the angle between them using the tangent difference formula. The extremum occurs when the derivative of tan(θ) with respect to the parameter equals zero.
<p><strong>Step 1:</strong> Let the common ordinate be the vertical line x = c where 0 < c < a. On this line, points are P(c, y₁) on ellipse and Q(c, y₂) on circle.</p><p><strong>Step 2:</strong> For ellipse: $\frac{c^2}{a^2} + \frac{y_1^2}{b^2} = 1$ gives $y_1 = \frac{b\sqrt{a^2-c^2}}{a}$</p><p>For circle: $c^2 + y_2^2 = a^2$ gives $y_2 = \sqrt{a^2-c^2}$</p><p><strong>Step 3:</strong> Tangent at P(c, y₁) to ellipse: slope $m_1 = -\frac{b^2 c}{a^2 y_1} = -\frac{bc}{a\sqrt{a^2-c^2}}$</p><p>Tangent at Q(c, y₂) to circle: slope $m_2 = -\frac{c}{y_2} = -\frac{c}{\sqrt{a^2-c^2}}$</p><p><strong>Step 4:</strong> Angle between tangents: $\tan\theta = \left|\frac{m_1-m_2}{1+m_1m_2}\right| = \left|\frac{c(1-b)}{a\sqrt{a^2-c^2} + bc^2}\right|$</p><p><strong>Step 5:</strong> Let $t = c^2$. To maximize $\tan\theta$, differentiate with respect to t and set = 0. This gives $t = a^2(a^2-b^2)/(a^2+b^2)$</p><p><strong>Step 6:</strong> Substituting back: $\tan\theta_{max} = \frac{a-b}{a+b}$</p><p>Therefore: $\theta_{max} = \tan^{-1}\left(\frac{a-b}{a+b}\right) = \frac{1}{2}\tan^{-1}\left(\frac{a^2-b^2}{ab}\right)$</p><p>Or equivalently: $\tan\left(2\theta\right) = \frac{a^2-b^2}{ab}$</p><p>∴ Answer: A</p>
Correct Answer: A