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Question

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


A
10
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B
210
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C
310
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D
410
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Solution

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

differentiate w.r.t x
$$2x+2y\dfrac{dy}{dx}+y+x\dfrac{dy}{dx}=0$$

$$\implies (x+2y)\dfrac{dy}{dx}=-(2x+y)\\\implies (\dfrac{dy}{dx})_{h,k}=-\dfrac{(2h+k)}{h+2k}$$

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

As this line also passes through (h,k)

$$k=\dfrac{h+2k}{2h+k}\times h\\\implies 2hk+k^2=h^2+2hk$$

we get $$k=\pm h$$

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

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

Physics

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