If A = {(x, y) : x^2 + y^2 = 25} and B = {(x, | vedclass

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

Question

(D) The set $A$ represents a circle with center $(0, 0)$ and radius $r = 5$,given by $x^2 + y^2 = 5^2$.The set $B$ represents an ellipse given by $x^2

Vedclass · Accepted Answer

Four points

Vedclass · Answer

(D) The set $A$ represents a circle with center $(0, 0)$ and radius $r = 5$,given by $x^2 + y^2 = 5^2$.The set $B$ represents an ellipse given by $x^2 + 9y^2 = 144$,which can be written as $\frac{x^2}{144} + \frac{y^2}{16} = 1$,or $\frac{x^2}{12^2} + \frac{y^2}{4^2} = 1$.Comparing the two,the circle has a radius of $5$. The ellipse has semi-major axis $a = 12$ along the $x$-axis and semi-minor axis $b = 4$ along the $y$-axis.Since the circle's radius $(5)$ is greater than the ellipse's semi-minor axis $(4)$ but less than its semi-major axis $(12)$,the circle intersects the ellipse at four distinct points.

If $A = \{(x, y) : x^2 + y^2 = 25\}$ and $B = \{(x, y) : x^2 + 9y^2 = 144\}$,then the number of points in $A \cap B$ is

Similar Questions

The locus of the point $P(x, y)$ satisfying the relation $\sqrt{(x - 3)^2 + (y - 1)^2} + \sqrt{(x + 3)^2 + (y - 1)^2} = 6$ is

If the points of intersection of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the circle $x^{2}+y^{2}=4b$ (where $b > 4$) lie on the curve $y^{2}=3x^{2}$,then $b$ is equal to:

The tangents drawn from the point $P(3, 4)$ to the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ touch the ellipse at points $A$ and $B$. The equation of the locus of a point which is equidistant from point $P$ and the line $AB$ is:

Equations of the latus rectum of the ellipse $9x^2+4y^2-18x-8y-23=0$ are:

If $4x + 2y + n = 0$ is a normal to the ellipse $\frac{x^2}{36} + \frac{y^2}{16} = 1$,then $n = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module