Question

The co-ordinate of a point on the auxiliary circle of the ellipse $x^{2}+2 y^{2}=4$ corresponding to the point on the ellipse whose eccentric angle is $60^{\circ}$ will be
(A) $(\sqrt{3}, 1)$
(B) $(1, \sqrt{3})$
(C) $(1,1)$
(D) $(1,2)$

Answer 1

Ans: (B)
Hint: $\frac{x^{2}}{4}+\frac{y^{2}}{2}=1$
Equation of auxiliary circle is $x^{2}+y^{2}=4$
$\therefore$ Point on the auxiliary circle is $p\left(2 \cos 60^{\circ}, 2 \sin 60^{\circ}\right)$ $p(1, \sqrt{3})$

Answer 2

buy tadalafil 20mg for sale <a href="https://ordergnonline.com/">tadalafil tablet</a> buy best erectile dysfunction pills

Answer 3

tadalafil 5mg over the counter <a href="https://ordergnonline.com/">order generic cialis 20mg</a> buy ed pills fda