Let an ellipse E:x2a2+y2b2=1, a2>b2, passes through 32,1 and has eccentricity 13 If a circle, centered at focus F(α, 0), α > 0, of E and radius 23 intersects E at two points P and Q, then PQ2 is equal to

e=13 b2=a2(1-e2)=a21-13 b2=2a23equation of ellipse passes through 32,1 is⇒32a2+1b2=1 ⇒32a2+32a2=1 ⇒2a2=6 ⇒ a2=3 ⇒b2=2 ∴ eq of an ellipse is x23+y22=1 centre of circle=Focus of ellipse=(1, 0)=(α, 0) radius=23=rCircle is (x-1)2+y2=43 point of intersection of x23+y22=1 and (x-1)2+y2=43x23+43-(x-1)22=1 2x2+4-3(x-1)2=6 2x2-3x2-3+6x=2 x2-6x+5=0 (x-1) (x-5)=0 x=1 or 5 y2=43(x-1)2 If x=1 then y2=43⇒y=±23 P1, 23 Q1, -23 PQ=43 ⇒PQ2=163

Let an ellipse E:x2a2+y2b2=1, a2>b2, passes through 32,1 and has eccentricity 13 If a circle, centered at focus F(α, 0), α > 0, of E and radius 23 intersects E at two points P and Q, then PQ2 is equal to

e=13 b2=a2(1-e2)=a21-13 b2=2a23equation of ellipse passes through 32,1 is⇒32a2+1b2=1 ⇒32a2+32a2=1 ⇒2a2=6 ⇒ a2=3 ⇒b2=2 ∴ eq of an ellipse is x23+y22=1 centre of circle=Focus of ellipse=(1, 0)=(α, 0) radius=23=rCircle is (x-1)2+y2=43 point of intersection of x23+y22=1 and (x-1)2+y2=43x23+43-(x-1)22=1 2x2+4-3(x-1)2=6 2x2-3x2-3+6x=2 x2-6x+5=0 (x-1) (x-5)=0 x=1 or 5 y2=43(x-1)2 If x=1 then y2=43⇒y=±23 P1, 23 Q1, -23 PQ=43 ⇒PQ2=163

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Let an ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a^{2} > b^{2},$ passes through $(\sqrt{\frac{3}{2}}, 1)$ and has eccentricity $\frac{1}{\sqrt{3}}$ If a circle, centered at focus $F (α, 0), α > 0,$ of E and radius $\frac{2}{\sqrt{3}}$ intersects E at two points P and Q, then PQ² is equal to

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$\frac{8}{3}$

$\frac{16}{3}$

$\frac{4}{3}$

answer is C.

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Detailed Solution

$e = \frac{1}{\sqrt{3}} b^{2} = a^{2} (1 - e^{2}) = a^{2} (1 - \frac{1}{3}) b^{2} = \frac{2 a^{2}}{3}$

equation of ellipse passes through $(\sqrt{\frac{3}{2}}, 1) i s$

$\Rightarrow \frac{3}{2 a^{2}} + \frac{1}{b^{2}} = 1 \Rightarrow \frac{3}{2 a^{2}} + \frac{3}{2 a^{2}} = 1 \Rightarrow 2 a^{2} = 6 \Rightarrow a^{2} = 3 \Rightarrow b^{2} = 2 ∴ eq of an ellipse i s \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 centre of circle = Focus of ellipse = (1, 0) = (α, 0) radius = \frac{2}{\sqrt{3}} = r$

${C i r c l e i s (x - 1)}^{2} + y^{2} = \frac{4}{3} point of intersection of \frac{x^{2}}{3} + \frac{y^{2}}{2} = 1 and {(x - 1)}^{2} + y^{2} = \frac{4}{3}$ $\frac{x^{2}}{3} + \frac{\frac{4}{3} - {(x - 1)}^{2}}{2} = 1 2 x^{2} + 4 - 3 {(x - 1)}^{2} = 6 2 x^{2} - 3 x^{2} - 3 + 6 x = 2 x^{2} - 6 x + 5 = 0 (x - 1) (x - 5) = 0 x = 1 o r 5 y^{2} = \frac{4}{3} {(x - 1)}^{2} I f x = 1 t h e n y^{2} = \frac{4}{3} \Rightarrow y = \pm \frac{2}{\sqrt{3}} P (1, \frac{2}{\sqrt{3}}) Q (1, - \frac{2}{\sqrt{3}}) PQ = \frac{4}{\sqrt{3}} \Rightarrow {PQ}^{2} = \frac{16}{3}$