Question

The point on the curve $4x^2+a^2{y}^2=4a^2,4<a^2<8,$ that is farthest from the point $(0,-2)$ is :

• A. $(2,0)$
• B. $(0,2)$
• C. $(2,-2)$
• D. $(-2,2)$

Answer 1

Answer: (B)

$(0,2)$

The given equation is $\frac{x^2}{a^2}+\frac{y^2}{4}=1,$ which represents an ellipse on which any point may be taken as $(a\cos \varphi ,2\sin \varphi )$
If $d$ be its distance from $(0,-2)$ then
let $z=d^2=a^2{\cos }^2\varphi +4(1+\sin \varphi {)}^2$
$\Rightarrow \frac{d_z}{d_{\varphi }}=-2a^2\cos \varphi \sin \varphi +8(1+\sin \varphi )\cos \varphi =0$ ....(i)
$=(4-a^2)]sin2\varphi +8\cos \varphi$
From (i) we get
$\cos \varphi =0$ or $\sin \varphi =\frac{4}{a^2-4}>1$ (rejected)
by given condition $4<a^2<8$
or $1<a^2/4<2$ and this value is rejected.
we choose $\cos \varphi =0\therefore \varphi =\frac{\pi }{2},$ so the point becomes $(0,2)$
Also $\frac{d^{2}z}{d{\varphi }^2}=(4-a^2)2\cos 2\varphi -8\sin \varphi$
$=(4-a^2)(-2)-8=2(a^2-8)=-ve$
as $4<a^2<8$ Hence $z=d^2$ is maximum.