In an ellipse 9{x^2} + 5{y^2} = 45 , distance between foci is

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10-2. Parabola, Ellipse, Hyperbola

easy

In an ellipse $9{x^2} + 5{y^2} = 45$, the distance between the foci is

$4\sqrt 5 $

$3\sqrt 5 $

$3$

$4$

Solution

(d) Given equation may be written as $\frac{{{x^2}}}{5} + \frac{{{y^2}}}{9} = 1.$

Comparing the given equation with standard equation, we get ${a^2} = 5$ and ${b^2} = 9.$

We also know that in an ellipse (where ${b^2} > {a^2})$ ${a^2} = {b^2}(1 – {e^2})\,\lambda $ or ${e^2} = \frac{{{b^2} – {a^2}}}{{{b^2}}} = \frac{{9 – 5}}{9} = \frac{4}{9}$

$e = \frac{2}{3}.$

Therefore distance between foci $ = 2be = 2 \times 3 \times \frac{2}{3} = 4$.

Standard 11

Mathematics

Number of points on the ellipse $\frac{{{x^2}}}{{50}} + \frac{{{y^2}}}{{20}} = 1$ from which pair of perpendicular tangents are drawn to the ellips $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{9}} = 1$

normal

The eccentricity of the ellipse $\frac{{{{(x – 1)}^2}}}{9} + \frac{{{{(y + 1)}^2}}}{{25}} = 1$ is

medium

The smallest possible positive slope of a line whose $y$-intercept is $5$ and which has a common point with the ellipse $9 x^2+16 y^2=144$ is

normal

(KVPY-2011)

The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ is :

hard

(JEE MAIN-2021)

On the ellipse $4{x^2} + 9{y^2} = 1$, the points at which the tangents are parallel to the line $8x = 9y$ are