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

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

Question

(d) The given ellipse is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$.

The value of the expression $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} - 1$ is po

Vedclass · Accepted Answer

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

Vedclass · Answer

(d) The given ellipse is $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$.

The value of the expression $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} - 1$ is positive for $x = 1,\,y = 2$ and negative for $x = 2,\,y = 1$.

Therefore $P$ lies outside $E$ and $Q$ lies inside $E$.

The value of the expression ${x^2} + {y^2} - 9$ is negative for both the points $P$ and $Q$.

Therefore $P$ and $Q$ both lie inside $C$.

Hence $P$ 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

The pole of the straight line $x + 4y = 4$ with respect to ellipse ${x^2} + 4{y^2} = 4$ is

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

For the ellipse $25{x^2} + 9{y^2} - 150x - 90y + 225 = 0$ the eccentricity $e = $

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 : . . . . .

Tangents at extremities of latus rectum of ellipse $3x^2 + 4y^2 = 12$ form a rhombus of area (in $sq.\ units$) -