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An ellipse is drawn by taking a diameter of the circle $(x-1)^2+y^2=4$ as its semi-minor axis and a diameter of the circle $x^2+(y-2)^2=16$ is a 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
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Verified Answer
The correct answer is:
$x^2+4 y^2=64$
Concept:
The equation of the circle with center at $(h, k)$ and radius $r$ is given by:
$(x-h)^2+(y-k)^2=r^2$
Calculation:
Equation of the circle is $(x-1)^2+y^2=4$, radius $=2$, diameter $=4$ Length of the semi-minor axis is 4
Equation of the circle is $x^2+(y-2)^2=16$, radius $=4$, diameter $=8$ Length of the semi-major axis is 8 , we know the equation of the ellipse
$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \frac{x^2}{(8)^2}+\frac{y^2}{(4)^2}=1 \\ & \frac{x^2}{64}+\frac{y^2}{16}=1 \\ & x^2+4 y^2=64\end{aligned}$
Hence, the equation of the ellipse is $x^2+4 y^2=64$.
The equation of the circle with center at $(h, k)$ and radius $r$ is given by:
$(x-h)^2+(y-k)^2=r^2$
Calculation:
Equation of the circle is $(x-1)^2+y^2=4$, radius $=2$, diameter $=4$ Length of the semi-minor axis is 4
Equation of the circle is $x^2+(y-2)^2=16$, radius $=4$, diameter $=8$ Length of the semi-major axis is 8 , we know the equation of the ellipse
$\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \\ & \frac{x^2}{(8)^2}+\frac{y^2}{(4)^2}=1 \\ & \frac{x^2}{64}+\frac{y^2}{16}=1 \\ & x^2+4 y^2=64\end{aligned}$
Hence, the equation of the ellipse is $x^2+4 y^2=64$.
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