If e is the eccentricity of the ellipse x2-a2-y2-b2-1 a - b- then

The correct option is B a^2=b^2(1 e^2) x ^ 2 a ^ 2 + y ^ 2 b ^ 2 =1 for (a gt;b) , #160; #160; #160; #160; b ^ 2 = a ^ 2 ( ...

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Byju's Answer

Standard XII

Mathematics

Logarithmic Function

If e is the...

Question

If $e$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(a < b)$ , then

b^{2} = a^{2} (1 - e^{2})

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a^{2} = b^{2} (1 - e^{2})

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a^{2} = b^{2} (e^{2} - 1)

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b^{2} = a^{2} (e^{2} - 1)

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Solution

The correct option is B $a^{2} = b^{2} (1 - e^{2})$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
for $(a > b)$ , $b^{2} = a^{2} (1 - e^{2})$
for $(a < b)$ , $a^{2} = b^{2} (1 - e^{2})$

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Similar questions

Given standard equation of ellipse,
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ ,
with eccentricity $e$ .
Match the following
$\begin{matrix} a) Major axis & i) 2 a (1 - e^{2}) b) Minor axis & i i) y = 0 c) Double ordinate & i i i) x = 0 d) Latus Rectum length & i v) x = \frac{- a}{e} v) \sqrt{1 - \frac{b^{2}}{a^{2}}} \end{matrix}$

Q. If the eccentricities of the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

and

\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1

e

and

e_{1}

, then

\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =

Q. Find the area bounded by the ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1

and the ordinates

x = 0

and

x = a e

, where

b^{2} = a^{2} (1 - e^{2})

and

e < 1

Q. If

e

is eccentricity of ellipse

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a > b)

and

e^{'}

is eccentricity of

\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 (a < b)

then

If the eccentricities of the hyperbolas $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 a n d \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} = 1$ be e and $e_{1}$ , then $\frac{1}{e^{2}} + \frac{1}{e_{1}^{2}} =$

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If e is the eccentricity of the ellipse x2a2+y2b2=1 (a<b), then

The correct option is B a2=b2(1−e2)x2a2+y2b2=1for (a>b) , b2=a2(1−e2)for (a<b), a2=b2(1−e2)

If $e$ is the eccentricity of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ $(a < b)$ , then

The correct option is B $a^{2} = b^{2} (1 - e^{2})$
$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
for $(a > b)$ , $b^{2} = a^{2} (1 - e^{2})$
for $(a < b)$ , $a^{2} = b^{2} (1 - e^{2})$