The equation of the ellipse with focus at ( ± 5 , 0 ) and x = 5 36 as one directrix is

We have ae = 5 [Since focus is ( ± ae , 0 ) ] and e a = 5 36 [ since directrix is x = ± e a ] On solving we get a = 6 and e = 6 5 ⇒ b 2 = a 2 ( 1 − e 2 ) = 36 ( 1 − 36 25 ) = 11 Thus, the required equation of the ellipse is 36 x 2 + 11 y 2 = 1

Mathematics - Ellipse Question with Solution | TestHub

MathematicsEllipseGeneralMedium2 minPYQ_2011

MathematicsMediumsingle choice

The equation of the ellipse with focus at $(\pm 5, 0)$ and $x = \frac{36}{5}$ as one directrix is

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We have ae $= 5$ [Since focus is $(\pm ae, 0)$ ] and $\frac{a}{e} = \frac{36}{5} [$ since directrix is $x = \pm \frac{a}{e}]$ On solving we get $a = 6$ $and e = \frac{5}{6} \Rightarrow b^{2} = a^{2} (1 - e^{2}) = 36 (1 - \frac{25}{36}) = 11$ Thus, the required equation of the ellipse is $\frac{x ^{2}}{36} + \frac{y ^{2}}{11} = 1$

Stream:BITSATSubject:MathematicsTopic:EllipseSubtopic:General

⏱ 2mℹ️ Source: PYQ_2011

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