The coordinate of a point on the auxiliary ci | vedclass

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(B) The given equation of the ellipse is $x^{2}+2y^{2}=4$. Dividing by $4$, we get $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$. Comparing this with the standa

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$(1, \sqrt{3})$

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(B) The given equation of the ellipse is $x^{2}+2y^{2}=4$. Dividing by $4$, we get $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$. Comparing this with the standard form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, we have $a^{2}=4$, so $a=2$. The auxiliary circle of an ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is given by $x^{2}+y^{2}=a^{2}$. Here, the equation of the auxiliary circle is $x^{2}+y^{2}=4$. The coordinates of a point on the auxiliary circle corresponding to an eccentric angle $	heta$ are given by $(a \cos 	heta, a \sin 	heta)$. Given $	heta = 60^{\circ}$ and $a=2$, the coordinates are $(2 \cos 60^{\circ}, 2 \sin 60^{\circ})$. Substituting the values, we get $(2 	imes \frac{1}{2}, 2 	imes \frac{\sqrt{3}}{2}) = (1, \sqrt{3})$. Thus, the correct option is $B$.

The coordinate of a point on the auxiliary circle of the ellipse $x^{2}+2y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be

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