An ellipse is drawn by taking a diameter of t | vedclass

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

Question

(D) The circle $(x - 1)^2 + y^2 = 1$ has a radius $r_1 = 1$,so its diameter is $2$. Given this is the semi-minor axis,$b = 2$.The circle $x^2 + (y - 2

Vedclass · Accepted Answer

$x^2 + 4y^2 = 16$

Vedclass · Answer

(D) The circle $(x - 1)^2 + y^2 = 1$ has a radius $r_1 = 1$,so its diameter is $2$. Given this is the semi-minor axis,$b = 2$.The circle $x^2 + (y - 2)^2 = 4$ has a radius $r_2 = 2$,so its diameter is $4$. Given this is the semi-major axis,$a = 4$.The standard equation of an ellipse centered at the origin with axes along the coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Substituting the values $a = 4$ and $b = 2$,we get $\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1$.$\Rightarrow \frac{x^2}{16} + \frac{y^2}{4} = 1$.Multiplying by $16$,we get $x^2 + 4y^2 = 16$.

An ellipse is drawn by taking a diameter of the circle $(x - 1)^2 + y^2 = 1$ as its semi-minor axis and a diameter of the circle $x^2 + (y - 2)^2 = 4$ as its semi-major axis. If the center of the ellipse is at the origin and its axes are the coordinate axes,then the equation of the ellipse is:

Similar Questions

Let the line $y=mx$ and the ellipse $2x^{2}+y^{2}=1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the coordinate axes at $(-\frac{1}{3\sqrt{2}}, 0)$ and $(0, \beta)$,then $\beta$ is equal to

$S=(-1, 1)$ is the focus,$2x-3y+1=0$ is the directrix corresponding to $S$,and $\frac{1}{2}$ is the eccentricity of an ellipse. If $(a, b)$ is the centre of the ellipse,then $3a+2b=$

The value of $k$,if $(1, 2)$ and $(k, -1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$,is

$A$ tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersects the coordinate axes at $A$ and $B$. Then the locus of the circumcentre of triangle $AOB$ (where $O$ is the origin) is:

If $C$ is the centre of the ellipse $9x^2 + 16y^2 = 144$ and $S$ is one focus,then the ratio of $CS$ to the major axis is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module