For a positive real number k let Ek be the ellipse with equation x2a2 - k - y2b2 - k - 1where a - b - 0- All members of the family of ellipse Ek - k - 0 have the same-

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. Find the equation of the ellipse having the latus recta of the ellipse

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

as tangents and the point

(0, \pm b)

as its focii.

Q. The locus of the point of intersection of the lines

b x t - a y t = a b

and

b x + a y = a b t

t

being parameter is

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For a positive real number k let Ek be the ellipse with equationx2a2+k+y2b2+k=1where a>b>0. All members of the family of ellipse {Ek:k>0} have the same.

The correct options are B Foci C CentreGiven ellipse is x2a2+k+x2b2+k=1,Eccentricity of the given ellipse is e2=b2−a2k+a2This depends on k,the focci are ±(ae,0)=±(√b2−a2,0),independent of k,∴ all the ellipse of this family have the same focci,centre is always origin (0,0).

For a positive real number $k$ let $E_{k}$ be the ellipse with equation
$\frac{x^{2}}{a^{2} + k} + \frac{y^{2}}{b^{2} + k} = 1$
where $a > b > 0$ . All members of the family of ellipse ${E_{k} : k > 0}$ have the same.