Question

The points on the curve $5x^2-6xy+5y^2-4=0$ that are the nearest the origin are

• A. $(1/2,-1/2),(-1/2,1/2)$
• B. $(0,2/\sqrt{5}),(0,-2/\sqrt{5})$
• C. $(2/\sqrt{5},0),(-2/\sqrt{5},0)$
• D. none of these

Answer 1

Answer: (A)

$(1/2,-1/2),(-1/2,1/2)$

Differentiating the given equation , we have

$10x-6y-6y{y}^{\prime }\left(x\right)+10y{y}^{\prime }\left(x\right)=0\Rightarrow y^{\prime }\left(x\right)=\frac{(3y-5x)}{(5y-3x)}$

The square of the distance of $(x,y)$ from origin is S

$=x^2+y^2=\frac{2(2+3xy)}{5}$

($(x,y)$ is a point on the given curve)

$\frac{dS}{dx}=\frac{6}{5}y+\frac{6}{5}x\frac{dy}{dx}=\frac{6}{5}[y+x\frac{(3y-5x)}{5y-3x}]=\frac{6}{5}\left[\frac{{5y}^2-{5x}^2}{5y-3x}\right]$

$S^{\prime }\left(x\right)=0\iff .y=\pm x$ if $y=x$ then

${5x}^2-6xy+{5y}^2=4\Rightarrow {10x}^2={6x}^2=4\Rightarrow x=\pm 1$

If $y=-x\Rightarrow {16x}^2=4\Rightarrow x=\pm 1/2$

Thus the extremum point are $(1,1),(-1,-1),(1/2-1/2),(-1/2-1/2)$