Question

Find the points on the curve $a{x}^2+2bxy+a{y}^2=c;c>b>a>0$, whose distance from the origin is minimum.

• A. $\left(\sqrt{\frac{c}{2(a+b)}},\sqrt{\frac{c}{2(a+b)}}\right)$
• B. $\left(\sqrt{\frac{c}{2(a+b)}},-\sqrt{\frac{c}{2(a+b)}}\right)$
• C. $\left(-\sqrt{\frac{c}{(a+b)}},-\sqrt{\frac{c}{(a+b)}}\right)$
• D. None of these

Answer 1

Answer: (A), (B)

$\left(\sqrt{\frac{c}{2(a+b)}},-\sqrt{\frac{c}{2(a+b)}}\right)$
$\left(\sqrt{\frac{c}{2(a+b)}},\sqrt{\frac{c}{2(a+b)}}\right)$

Differentiating with respect to x, we get
$2ax+2ay.y^{\prime }+2by+2bx{y}^{\prime }=0$
Or
$ax+by+y^{\prime }(ay+bx)=0$
Or
$y^{\prime }=\frac{-(ax+by)}{ay+bx}$
Hence slope of normal
$\frac{-1}{y^{\prime }}=\frac{ay+bx}{ax+by}$
Now distance from origin of a point lying on the above curve is minimum when it lies on the normal.
Let the point be $x_1,y_1$
Slope of the line will be $=\frac{y_1}{x_1}$ ...(joining the origin and the point).
Hence
$\frac{a{y}_1+b{x}_1}{a{x}_1+b{y}_1}=\frac{y_1}{x_1}$
Or
$a{y}_1.x_1+b{x}_1^2=a{x}_1{y}_1+b{y}_1^2$
Or
$x_1=\pm y_1$.
Now if $x_1=y_1=k$,
we get
$a{k}^2+2b{k}^2+a{k}^2=c$
Or
$2k^2(a+b)=c$
Or
$x=y=\sqrt{\frac{c}{2(a+b)}}$.
If $x=-y$ we get
$a{x}^2-2b{x}^2+a{x}^2=c$
$2x^2(a-b)=c$
Or
$x=\sqrt{\frac{c}{2(a-b)}}$.
And
$y=-\sqrt{\frac{c}{2(a-b)}}$