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 Δ PF1F2 lies on the ellipse, where F1 and F2 are the foci of the ellipse, then the square of the eccentricity of the ellipse is

Question

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 Δ PF1F2 lies on the ellipse, where F1 and F2 are the foci of the ellipse, then the square of the eccentricity of the ellipse is (A) (√5 - 1/2) (B) √3-1 (C) (1/√2) (D) (√3/2)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-spnoo9hndvwd.png" /> mF1 B⋅ mP F2=-1⇒ (a - 0/0 - a e)× (b - 0/0 - (- a e))=-1⇒ (b2/a2)=e2⇒ e4+e2-1=0 ∴ e2=((√5 - 1/2))

$\frac{5 - 1}{2}$

$3 - 1$

$\frac{1}{2}$

$\frac{3}{2}$

$m_{F_{1} B} \cdot m_{P F_{2}} = - 1 \Rightarrow \frac{a - 0}{0 - a e} \times \frac{b - 0}{0 - ( - a e )} = - 1 \Rightarrow \frac{b ^{2}}{a ^{2}} = e^{2} \Rightarrow e^{4} + e^{2} - 1 = 0$
$∴ e^{2} = (\frac{5 - 1}{2})$

Q. If the circle whose diameter is the major axis of the ellipse a2x2​+b2y2​=1(a>b>0) meets the minor axis at point P and the orthocentre of ΔPF1​F2​ lies on the ellipse, where F1​ and F2​ are the foci of the ellipse, then the square of the eccentricity of the ellipse is

25​−1​

3​−1

2​1​

23​​

mF1​B​⋅mPF2​​=−1⇒0−aea−0​×0−(−ae)b−0​=−1⇒a2b2​=e2⇒e4+e2−1=0 ∴e2=(25​−1​)

$\frac{5 - 1}{2}$

$3 - 1$

$\frac{1}{2}$

$\frac{3}{2}$

$m_{F_{1} B} \cdot m_{P F_{2}} = - 1 \Rightarrow \frac{a - 0}{0 - a e} \times \frac{b - 0}{0 - ( - a e )} = - 1 \Rightarrow \frac{b ^{2}}{a ^{2}} = e^{2} \Rightarrow e^{4} + e^{2} - 1 = 0$
$∴ e^{2} = (\frac{5 - 1}{2})$