Match the following for ellipse- x2-a2-y2-b2-1- a-b with eccentricity e-

The correct option is Let's deduce what all we can infer from x^2/a^2+y^2/b^2=1 Foci are (ae,0),( ae,0) Directrices are x=±a/e Eccentricity, e is √(1 b^2/a^2) V ...

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Standard XI

Mathematics

Terminologies Related to an Ellipse

Match the fol...

Question

Match the following for ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ with eccentricity $e$ .

(\pm a e, 0)

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x = \pm \frac{a}{e}

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(\pm a, 0)

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1-b2a2

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Solution

Let's deduce what all we can infer from $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Foci are $(a e, 0), (- a e, 0)$
Directrices are $x = \pm \frac{a}{e}$
Eccentricity, $e$ is $\sqrt{1 - \frac{b^{2}}{a^{2}}}$
Vertices are $(a, 0), (- a, 0)$

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Similar questions

Q. Match the following for ellipse,

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

, with eccentricity

e .

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

Q. If

e

is the eccentricity of the ellipse

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

(a < b)

, then

Q. In an ellipse,

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

, where a < b, eccentricity 'e' is given by

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

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Match the following for ellipse, x2a2+y2b2=1, a>b with eccentricity e.

Let's deduce what all we can infer from x2a2+y2b2=1 Foci are (ae,0),(−ae,0) Directrices are x=±ae Eccentricity, e is √1−b2a2 Vertices are (a, 0), (−a, 0)

Match the following for ellipse, $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b$ with eccentricity $e$ .

Let's deduce what all we can infer from $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$
Foci are $(a e, 0), (- a e, 0)$
Directrices are $x = \pm \frac{a}{e}$
Eccentricity, $e$ is $\sqrt{1 - \frac{b^{2}}{a^{2}}}$
Vertices are $(a, 0), (- a, 0)$