An ellipse with foci - 3-0 passes through 4-1 then its eccentricity is

The correct option is C 1/√(2) Distance #160;between #160;foci =2ae #160; =6. 2a = SP+SP^' = √((4 3)^2+ (1)^2) + √((4+3)^2 + 1 ...

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Standard XII

Mathematics

Definition of Functions

An ellipse wi...

Question

An ellipse with foci $(\pm 3, 0)$ passes through $(4, 1)$ then its eccentricity is

\frac{1}{2}

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\frac{1}{3}

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\frac{2}{3}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is C $\frac{1}{\sqrt{2}}$
Distance between foci $= 2 a e$ $= 6.$
$2 a = S P + S P^{^{'}}$
$= \sqrt{(4 - 3)^{2} + (1)^{2}} + \sqrt{(4 + 3)^{2} + 1}$
$= \sqrt{2} + \sqrt{50}$
$= 6 \sqrt{2}$
$∴ e = \frac{1}{\sqrt{2}}$

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An ellipse with foci (±3,0) passes through (4,1) then its eccentricity is

The correct option is C 1√2Distance between foci =2ae =6.2a=SP+SP′=√(4−3)2+(1)2+√(4+3)2+1=√2+√50=6√2∴e=1√2

An ellipse with foci $(\pm 3, 0)$ passes through $(4, 1)$ then its eccentricity is

The correct option is C $\frac{1}{\sqrt{2}}$
Distance between foci $= 2 a e$ $= 6.$
$2 a = S P + S P^{^{'}}$
$= \sqrt{(4 - 3)^{2} + (1)^{2}} + \sqrt{(4 + 3)^{2} + 1}$
$= \sqrt{2} + \sqrt{50}$
$= 6 \sqrt{2}$
$∴ e = \frac{1}{\sqrt{2}}$