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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Ellipse

NDA 2019 · Mathematics · Ellipse

If the angle between the lines joining the end points of minor axis...

If the angle between the lines joining the end points of minor axis of the ellipse x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 with one of its foci is π2\frac{\pi}{2}, then what is the eccentricity of the ellipse?

Options

  1. A12\frac{1}{2}
  2. B12\frac{1}{\sqrt{2}}
  3. C32\frac{\sqrt{3}}{2}
  4. D122\frac{1}{2 \sqrt{2}}

Correct answer

B. 12\frac{1}{\sqrt{2}}

Step-by-step solution

Given, x2a2+y2b2=1\frac{x^2}{a^2}+\frac{y^2}{b^2}=1
given: ∠PF1Q=90∘\angle P F_1 Q=90^{{}^{\circ}}∴mPF1×mQF1=−1⇒−bae×bae=−1⇒b2=a2⋅e2⋯(i)∵b2=a2(1−e2)...(ii)\begin{array}{ll} \therefore & m_{P F_1} \times m_{Q F_1}=-1 \\ \Rightarrow & \frac{-b}{a e} \times \frac{b}{a e}=-1 \\ \Rightarrow & b^2=a^2 \cdot e^2 \cdots(i)\\ \because & b^2=a^2\left(1-e^2\right)...(ii) \end{array}From(i)and(ii)weget,From (i) and (ii) we get,e2=1−e2⇒2e2=1⇒e2=12⇒e=12\begin{aligned}& e²=1-e² \\ & \Rightarrow 2 e²=1 \\ & \Rightarrow e²=\frac{1}{2}\\ & \Rightarrow e=\frac{1}{\sqrt{2}} \end{aligned}$$
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