Tangents are drawn from points oncircle x^2 + y^2 = 49 to ellipse frac{{{x^2}}}{{25}} + frac{{

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

normal

Tangents are drawn from points onthe circle $x^2 + y^2 = 49$ to the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{24}} = 1$ angle between the tangents is

A

$\frac{\pi }{4}$

B

$\frac{\pi }{2}$

C

$\frac{\pi }{3}$

D

$\frac{\pi }{8}$

Solution

$x^{2}+y^{2}=49$ is director circle of $\frac{x^{2}}{25}+\frac{y^{2}}{24}=1$

Standard 11

Mathematics

Share

Similar Questions

The angle between the pair of tangents drawn to the ellipse $3{x^2} + 2{y^2} = 5$ from the point $(1, 2)$, is

hard

Let $E$ be the ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$ and $C$ be the circle ${x^2} + {y^2} = 9$. Let $P$ and $Q$ be the points $(1, 2)$ and $(2, 1)$ respectively. Then

easy

(IIT-1994)

Let the ellipse, $E _1: \frac{ x ^2}{ a ^2}+\frac{ y ^2}{b^2}=1, a > b$ and $E _2: \frac{ x ^2}{A^2}+\frac{ y ^2}{B^2}=1, A< B$ have same eccentricity $\frac{1}{\sqrt{3}}$. Let the product of their lengths of latus rectums be $\frac{32}{\sqrt{3}}$, and the distance between the foci of $E_1$ be $4$. If $E_1$ and $E_2$ meet at $A, B, C$ and $D$, then the area of the quadrilateral $A B C D$ equals:

hard

(JEE MAIN-2025)

If the line $y = 2x + c$ be a tangent to the ellipse $\frac{{{x^2}}}{8} + \frac{{{y^2}}}{4} = 1$, then $c = $

medium

Two sets $A$ and $B$ are as under:

$A = \{ \left( {a,b} \right) \in R \times R:\left| {a – 5} \right| < 1 \,\,and\,\,\left| {b – 5} \right| < 1\} $; $B = \left\{ {\left( {a,b} \right) \in R \times R:4{{\left( {a – 6} \right)}^2} + 9{{\left( {b – 5} \right)}^2} \le 36} \right\}$ then : . . . . .

hard

(JEE MAIN-2018)