Question

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

• A. $(-1,0)$
• B. $(0,2)$
• C. $(1,0)$
• D. $(\frac{\sqrt{3}}{2}a,1)$

Answer 1

Answer: (B)

$(0,2)$

Given equation of curve is

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

$\frac{x^2}{a^2}+\frac{y^2}{4}=1$

which represents an ellipse .

Let $P(a\cos \varphi ,2\sin \varphi )$ be any point on the ellipse.

Let $d$ be its distance from $(0,-2)$

Let $z=d^2=a^2{\cos }^2\varphi +4{(1+\sin \varphi )}^2$

$\frac{dz}{d\varphi }=-2a^2\cos \varphi \sin \varphi +8(1+\sin \varphi )\cos \varphi$

$\frac{dz}{d\varphi }=(4-a^2)\sin 2\varphi +8\cos \varphi$

$\Rightarrow \frac{dz}{d\varphi }=2\cos \varphi (-a^2\sin \varphi +4+4\sin \varphi )$

For maximum or minimum,

$\frac{dz}{d\varphi }=0$

$\Rightarrow 2\cos \varphi (-a^2\sin \varphi +4+4\sin \varphi )=0$

$\Rightarrow \cos \varphi =0$ or $\sin \varphi =\frac{4}{a^2-4}=\frac{1}{\frac{a^2}{4}-1}>1$ (rejected) (As given $4<a^2<8$ or $1<\frac{a^2}{4}<2$

So, we have $\cos \varphi =0$

$\therefore \varphi =\frac{\pi }{2}$

So, the point becomes $(0,2).$

Also $\frac{d^{2}z}{d{\varphi }^2}=(4-a^2)2\cos 2\varphi -8\sin \varphi .$

$=(4-a^2)(-2)-8=2(a^2-8)=-ive,$ as $4<a^2<8$

Hence $z=d^2$ is maximum.