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 centre of the ellipse is at the origin and its axes are co ordinate axes then, the ellipse is

4 x^{2} + y^{2} = 4

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x^{2} + 4 y^{2} = 8

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4 x^{2} + y^{2} = 16

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x^{2} + 4 y^{2} = 16

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Solution

The correct option is C $4 x^{2} + y^{2} = 16$
$e q 1 : (x - 1)^{2} + y^{2} = 1$
Diameter of this circle $= 2$
$e q 2 : x^{2} + (y - 2)^{2} = 4$
Diameter of this circle $= 4$

length of semi-minor axis $= 2$ [along X-axis]
length of semi major axis $= 4$ [along Y-axis]
required ellipse $e q 3 : \frac{x^{2}}{2^{2}} + \frac{y^{2}}{4^{2}} = 1 \Rightarrow 4 x^{2} + y^{2} = 16$

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An ellipse is drawn by taking a diameter of the circle (x−1)2+y2=1 as its semi minor axis and a diameter of the circle x2+(y−2)2=4 as its semi-major axis. If the centre of the ellipse is at the origin and its axes are co ordinate axes then, the ellipse is

The correct option is C 4x2+y2=16eq1:(x−1)2+y2=1Diameter of this circle =2eq2:x2+(y−2)2=4Diameter of this circle =4length of semi-minor axis =2 [along X-axis]length of semi major axis =4 [along Y-axis]required ellipse eq3:x222+y242=1⇒4x2+y2=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 centre of the ellipse is at the origin and its axes are co ordinate axes then, the ellipse is