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. $(0,0)$
• B. $(0,1)$
• C. $(0,2)$
• D. $(0,3)$

Answer 1

Answer: (C)

$(0,2)$

From the above equation of curve, we get
$y=\frac{2}{a}\sqrt{a^2-x^2}$ ...(i)
Now let the point be $P=(x,y)$
$=(x,\frac{2}{a}\sqrt{a^2-x^2})$
Then distance between $P$ and $(0,-2)$ will be
$D^2=x^2+(\frac{2}{a}\sqrt{a^2-x^2}+2{)}^2$
Differentiating the above expression with respect to x, we get
$2x+2(\frac{2}{a}\sqrt{a^2-x^2}+2).\frac{2}{a}.\frac{-2x}{2\sqrt{a^2-x^2}}$
$=2[x-(\frac{2}{a}\sqrt{a^2-x^2}+2).\frac{2x}{a\sqrt{a^2-x^2}}]$
$=2x[1-(\frac{2}{a}\sqrt{a^2-x^2}+2).\frac{2}{a\sqrt{a^2-x^2}}]$
$=0$
Then
$x=0$ or
$1-(\frac{2}{a}\sqrt{a^2-x^2}+2).\frac{2}{a\sqrt{a^2-x^2}}=0$
Considering $x=0$ we substitute in i, we get
$y=2$.
Hence one point is $(0,2)$.