An ellipse is drawn such that the diameter of | vedclass

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

Question

(A) The given circle is $(x - 1)^2 + y^2 = 1$. Its radius is $r_1 = 1$,so its diameter is $d_1 = 2$. This is the semi-minor axis length,$b = 2$.The se

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given circle is $(x - 1)^2 + y^2 = 1$. Its radius is $r_1 = 1$,so its diameter is $d_1 = 2$. This is the semi-minor axis length,$b = 2$.The second circle is $x^2 + (y - 2)^2 = 4$. Its radius is $r_2 = 2$,so its diameter is $d_2 = 4$. This is the semi-major axis length,$a = 4$.The standard equation of an ellipse with center at origin and axes as coordinate axes is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (assuming major axis along $x$-axis) or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (assuming major axis along $y$-axis).Case $1$: If major axis is along $y$-axis,$a = 4$ and $b = 2$. The equation is $\frac{x^2}{2^2} + \frac{y^2}{4^2} = 1$ $\Rightarrow \frac{x^2}{4} + \frac{y^2}{16} = 1$ $\Rightarrow 4x^2 + y^2 = 16$.Case $2$: If major axis is along $x$-axis,$a = 4$ and $b = 2$. The equation is $\frac{x^2}{4^2} + \frac{y^2}{2^2} = 1$ $\Rightarrow \frac{x^2}{16} + \frac{y^2}{4} = 1$ $\Rightarrow x^2 + 4y^2 = 16$.Comparing with the given options,$4x^2 + y^2 = 16$ is present in option $A$.

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

Similar Questions

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$. Then the maximum perpendicular distance of $P$ from the line $x - y = 10$ is (in $\sqrt{2}$)

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of its major axis. Its eccentricity is:

The eccentricity of the ellipse $x^2+4 y^2+2 x+16 y+13=0$ is

If the normal at one end of a latus rectum of the ellipse $\frac{x^2}{32}+\frac{y^2}{b^2}=1$ passes through one end of the minor axis,then $\frac{e^4}{1-e^2}=$ (Here $e$ is the eccentricity of the ellipse)

If $\frac{\pi}{3}$ and $\theta$ are the eccentric angles of the ends of a focal chord of the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$,then $\tan \theta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module