If the circle whose diameter is the major axis of the ellipse (x2)/(a2) + (y2)/(b2) = 1 (a > b > 0) meets the minor axis at point P and the orthocentre of ∆ P F _ 1 F _ 2 lies on the ellipse, where F _ 1 and F _ 2 are

Correct answer: A. 5 - 1 2 m _ F _ 1 B ⋅ m _ P F _ 2 = - 1 ⇒ (a - 0)/(0 - ae) × b - 0 0 - (- ae ) = - 1 ⇒ (b2)/(a2) = e^ 2 ⇒ e^ 4 + e^ 2 - 1 = 0 ∴ e^ 2 = ( 5 - 1 2 )

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Ellipse

NTA Abhyas 2020 · Mathematics · Ellipse

If the circle whose diameter is the major axis of the ellipse...

If the circle whose diameter is the major axis of the ellipse

\dfrac{x2}{a2} + \dfrac{y2}{b2} = 1

\left(a > b > 0\right)

meets the minor axis at point

and the orthocentre of

∆ P{F}_{1}{F}_{2}

lies on the ellipse, where

{F}_{1}

and

{F}_{2} are the foci of the ellipse, then the square of the eccentricity of the ellipse is If the circle whose diameter is the major axis of the ellipse (x2)/(a2) + (y2)/(b2) = 1 (a > b > 0) meets the minor

If the circle whose diameter is the major axis of the ellipse (x2)/(a2) + (y2)/(b2) = 1 (a > b > 0) meets the minor

Options

A $\dfrac{\sqrt{5} - 1}{2}$
B $\sqrt{3} - 1$
C $\dfrac{1}{\sqrt{2}}$
D $\dfrac{\sqrt{3}}{2}$

Correct answer

A. $\dfrac{\sqrt{5} - 1}{2}$

Step-by-step solution

If the circle whose diameter is the major axis of the ellipse (x2)/(a2) + (y2)/(b2) = 1 (a > b > 0) meets the minor

${m}_{{F}_{1}B} ⋅ {m}_{P{F}_{2}} = - 1 ⇒ \dfrac{a - 0}{0 - ae} \times \dfrac{b - 0}{0 - \left(-\text{ ae }\right)}$ = - 1 ⇒ \dfrac{b2}{a2} = e^{2} ⇒ e^{4} + e^{2} - 1 = 0 $<br />$ ∴ e^{2} = $\left(\dfrac{\sqrt{5} - 1}{2}\right)</p>$

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