Let E be the ellipse frac{x^2}{9} + frac{y^2} | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The ellipse $E$ is given by $f(x, y) = \frac{x^2}{9} + \frac{y^2}{4} - 1 = 0$. For point $P(1, 2)$,$f(1, 2) = \frac{1}{9} + \frac{4}{4} - 1 = \fra

Vedclass · Accepted Answer

$P$ lies inside $C$ but outside $E$

Vedclass · Answer

(D) The ellipse $E$ is given by $f(x, y) = \frac{x^2}{9} + \frac{y^2}{4} - 1 = 0$. For point $P(1, 2)$,$f(1, 2) = \frac{1}{9} + \frac{4}{4} - 1 = \frac{1}{9} > 0$,so $P$ lies outside $E$. For point $Q(2, 1)$,$f(2, 1) = \frac{4}{9} + \frac{1}{4} - 1 = \frac{16+9-36}{36} = -\frac{11}{36} The circle $C$ is given by $g(x, y) = x^2 + y^2 - 9 = 0$. For point $P(1, 2)$,$g(1, 2) = 1^2 + 2^2 - 9 = 1 + 4 - 9 = -4 For point $Q(2, 1)$,$g(2, 1) = 2^2 + 1^2 - 9 = 4 + 1 - 9 = -4 Thus,$P$ lies inside $C$ but outside $E$.

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

Similar Questions

If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

The eccentricity of the ellipse $9x^2 + 5y^2 - 30y = 0$ is

The sides of the rectangle of greatest area that can be inscribed in the ellipse $x^2+4y^2=64$ are :

The eccentricity of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ is:

If $\tan \theta_1 \cdot \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at:

Vedclass Test Series

Exam Paper Generator

Online Exam Module