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 equation of the ellipse is $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and the circle is $C: x^2 + y^2 = 9$.For point $P(1, 2)$:Check for circle $C

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the ellipse is $E: \frac{x^2}{9} + \frac{y^2}{4} = 1$ and the circle is $C: x^2 + y^2 = 9$.For point $P(1, 2)$:Check for circle $C$: $1^2 + 2^2 = 1 + 4 = 5 Check for ellipse $E$: $\frac{1^2}{9} + \frac{2^2}{4} = \frac{1}{9} + 1 = \frac{10}{9} > 1$,so $P$ is outside $E$.For point $Q(2, 1)$:Check for circle $C$: $2^2 + 1^2 = 4 + 1 = 5 Check for ellipse $E$: $\frac{2^2}{9} + \frac{1^2}{4} = \frac{4}{9} + \frac{1}{4} = \frac{16+9}{36} = \frac{25}{36} Thus,$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 in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

During electro-osmosis of $Fe(OH)_3$ sol,

Which of the following is the most basic oxide?

If $BaCl_2$ ionizes to an extent of $80 \ \%$ in aqueous solution,the value of van't Hoff factor is

The angle between the lines whose direction cosines are $\left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2}\right)$ and $\left(\frac{\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2}\right)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module