Shortest distance between $(0, 0)$ and $x^{2} + y^{2} + x y = 60$ is

\sqrt{10}

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2 \sqrt{10}

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3 \sqrt{10}

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4 \sqrt{10}

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Solution

The correct option is B $2 \sqrt{10}$
Given curve is $x^{2} + y^{2} + x y = 60$

differentiate w.r.t x
$2 x + 2 y \frac{d y}{d x} + y + x \frac{d y}{d x} = 0$

$⟹ (x + 2 y) \frac{d y}{d x} = - (2 x + y) ⟹ (\frac{d y}{d x})_{h, k} = - \frac{(2 h + k)}{h + 2 k}$

Let the line from (0,0) cuts the curve at (h,k)
Equation of line is
$y - 0 = \frac{h + 2 k}{2 h + k} (x - 0)$

As this line also passes through (h,k)

$k = \frac{h + 2 k}{2 h + k} \times h ⟹ 2 h k + k^{2} = h^{2} + 2 h k$

we get $k = \pm h$

Since curve is also passes through (h,k)
so
$h^{2} + k^{2} + h k = 60 ⟹ h^{2} + h^{2} + h^{2} = 60 ⟹ h^{2} = 20 ⟹ h = \sqrt{20}$

Distance of (0,0) from (h,k)= $\sqrt{20 + 20} = \sqrt{40} = 2 \sqrt{10}$

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Shortest distance between (0,0) and x2+y2+xy=60 is

Shortest distance between $(0, 0)$ and $x^{2} + y^{2} + x y = 60$ is