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Question: If P is a point of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, whose focii are S and S'. Le...

If P is a point of the ellipse x2a2+y2b2=1\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, whose focii are S and S'. Let ∠PSS′=α\angle PSS' = \alpha and ∠PS′S=β\angle PS'S = \beta, then

A

PS + PS' = 2a, if a > b

B

PS + PS' = 2b, if a < b

C

tanα2tanβ2=1−e1+etan \frac{\alpha}{2} tan \frac{\beta}{2} = \frac{1-e}{1+e}

D

tanα2tanβ2=a2−b2b2[a−a2−b2]tan \frac{\alpha}{2} tan \frac{\beta}{2} = \frac{\sqrt{a^2-b^2}}{b^2} [a-\sqrt{a^2-b^2}] when a > b

Answer

A, B, C

Explanation

Solution

The definition of an ellipse states that for any point P on the ellipse, the sum of the distances from P to the two foci (S and S') is constant. This constant sum is equal to 2a2a if the major axis is horizontal (length 2a2a) and 2b2b if the major axis is vertical (length 2b2b). Option (A) states that PS+PS′=2aPS + PS' = 2a if a>ba > b. This is true when the major axis is along the x-axis, with length 2a2a. Option (B) states that PS+PS′=2bPS + PS' = 2b if a<ba < b. This is true when the major axis is along the y-axis, with length 2b2b. Both statements are correct definitions of the sum of focal distances for an ellipse depending on its orientation.

For any ellipse, let ee be its eccentricity. For a point P on the ellipse, with foci S and S', let α=∠PSS′\alpha = \angle PSS' and β=∠PS′S\beta = \angle PS'S. It can be shown that tan⁡(α/2)tan⁡(β/2)=1−e1+e\tan(\alpha/2) \tan(\beta/2) = \frac{1-e}{1+e}. This formula is independent of the orientation of the ellipse. Thus, option (C) is correct.

Option (D) provides an expression for tan⁡(α/2)tan⁡(β/2)\tan(\alpha/2) \tan(\beta/2) when a>ba > b. When a>ba > b, the eccentricity is e=a2−b2ae = \frac{\sqrt{a^2-b^2}}{a}. Substituting this into 1−e1+e\frac{1-e}{1+e} gives 1−a2−b2a1+a2−b2a=a−a2−b2a+a2−b2\frac{1 - \frac{\sqrt{a^2-b^2}}{a}}{1 + \frac{\sqrt{a^2-b^2}}{a}} = \frac{a - \sqrt{a^2-b^2}}{a + \sqrt{a^2-b^2}}. Rationalizing this expression yields (a−a2−b2)2b2\frac{(a - \sqrt{a^2-b^2})^2}{b^2}. Option (D) states a2−b2b2[a−a2−b2]\frac{\sqrt{a^2-b^2}}{b^2} [a-\sqrt{a^2-b^2}]. For these to be equal, (a−a2−b2)2=a2−b2[a−a2−b2](a - \sqrt{a^2-b^2})^2 = \sqrt{a^2-b^2} [a-\sqrt{a^2-b^2}], which implies either a−a2−b2=0a - \sqrt{a^2-b^2} = 0 (meaning b=0b=0, not an ellipse) or a−a2−b2=a2−b2a - \sqrt{a^2-b^2} = \sqrt{a^2-b^2} (meaning a=2a2−b2a = 2\sqrt{a^2-b^2}, or 3a2=4b23a^2=4b^2). Since this equality does not hold for all a>ba > b, option (D) is incorrect.