Radius of circle having its centre at (0, 3) and passing through foci of ellipse frac{{{x^2}}}{{16}}

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

medium

The radius of the circle having its centre at $(0, 3)$ and passing through the foci of the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$, is

A

$3$

B

$3.5$

C

$4$

D

$\sqrt {12} $

(IIT-1995)

Solution

(c) The co-ordinates of foci are $( \pm ae,\,0)$. Here $a = 4,\,\,b = 3$

$\therefore$ ${b^2} = {a^2}(1 – {e^2})$ ==> $9 = 16(1 – {e^2})$ ==> $\frac{9}{{16}} = 1 – {e^2}$

==> $e = \pm \sqrt {\frac{7}{4}} $; $\therefore$ Points are $( \pm \sqrt 7 ,0)$.

$\therefore$ Radius $ = \sqrt {{{(\sqrt 7 – 0)}^2} + {{(0 – 3)}^2}} = \sqrt {7 + 9} = \sqrt {16} = 4$.

Standard 11

Mathematics

Share

Similar Questions

If the distance between a focus and corresponding directrix of an ellipse be $8$ and the eccentricity be $1/2$, then length of the minor axis is

easy

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 ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is :-

normal

The locus of the point of intersection of mutually perpendicular tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$, is

normal

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is

hard

(KVPY-2020)

An ellipse, with foci at $(0, 2)$ and $(0, -2)$ and minor axis of length $4$, passes through which of the following points?

hard

(JEE MAIN-2019)