Pθ1 and Qθ2 are two points on the ellipse x2a2+y2b2=1 with eccentricity e. If PSQ is a focal chord and tanθ12tanθ22=-22+3, then e and S are

TS EAMCET · Maths · Ellipse

$P (θ_{1})$ and $Q (θ_{2})$ are two points on the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ with eccentricity $e$ . If $P S Q$ is a focal chord and $\tan (\frac{θ_{1}}{2}) \tan (\frac{θ_{2}}{2}) = - (2 \sqrt{2} + 3)$ , then $e$ and $S$ are

A $\frac{1}{\sqrt{3}}, (\frac{a}{\sqrt{3}}, 0)$
B $\frac{1}{\sqrt{3}}, (\frac{- a}{\sqrt{3}}, 0)$
C $\frac{1}{\sqrt{2}}, (\frac{a}{\sqrt{2}}, 0)$
D $\frac{1}{\sqrt{2}}, (\frac{- a}{\sqrt{2}}, 0)$

Verified Solution

Answer & Solution

Correct Answer

(D) $\frac{1}{\sqrt{2}}, (\frac{- a}{\sqrt{2}}, 0)$

Step-by-step Solution

Detailed explanation

Here, S≡ae,0. Let Pacosθ1,bsinθ1 and Qacosθ2,bsinθ2 are two points on the ellipse x2a2+y2b2=1, so Slope of PS=Slope of QS ⇒0-bsinθ1ae-acosθ1=0-bsinθ2ae-acosθ2 ⇒sinθ1e-cosθ1=sinθ2e-cosθ2…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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