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Problem :

Find the number of common tangents to $y^2=2012x$ and $xy =(2013)^2$

Solution : Common tangent will have slope equal to both curves. therefore, differentiation both the curves we get the slopes .

$\therefore 2y\frac{dy}{dx}=2012 \Rightarrow \frac{dy}{dx} = \frac{2012}{2y} .....(1)$ and $x\frac{dy}{dx}+y=0$ $\Rightarrow \frac{dy}{dx}=-\frac{y}{x}.....(2)$

(1) and (2) represent the slope of the same tangent therefore, 1 =2

$\Rightarrow \frac{2012 }{2y } = \frac{-y}{x} $ But I am not getting anything here,

So, I used the second method : Solving for both the curves we will get point of intersection,

$y^2=2012x \Rightarrow x =\frac{y^2}{2012}.....(1) $ $y =\frac{(2013)^2}{x}.......(2)$

Now putting value of y from (2) in equation (1) we get

$\frac{\frac{(2013)^4}{x^2}}{2012}=x$ But I think this will also not give any solution please suggest thanks.

Sachin
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1 Answers1

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Let $(s,t)$ be the point on $y^2=2012x$ and let $(u,v)$ be the point on $xy=(2013)^2$.

Then, we have $$t^2=2012s,\tag1$$ $$uv=(2013)^2.\tag2$$

Since $$y^2=2012x\Rightarrow 2y\cdot\frac{dy}{dx}=2012,$$ we know that the $y$-axis is tangent to this curve at the origin. However, since $y$-axis is not tangent to the curve $xy=(2013)^2$, we may assume that $t\not=0$. Then, we have $$\frac{dy}{dx}=\frac{2012}{2y}.\tag3$$

On the other hand, since $u\not=0$, we have $$xy=(2013)^2\Rightarrow y+x\cdot\frac{dy}{dx}=0\Rightarrow \frac{dy}{dx}=-\frac yx.\tag4$$ From $(1),(2),(3),(4)$, the tangent line both at $(s,t)$ and at $(u,v)$ can be represented as $$y-t=\frac{2012}{2t}(x-s)\iff y=\frac{2012}{2t}x+\frac{2t^2-2012s}{2t}=\frac{2012}{2t}x+\frac t2$$ and $$y-v=-\frac{v}{u}(x-u)\iff y=-\frac{v^2}{(2013)^2}x+2v$$ respectively.

Hence, we have $$\frac{2012}{2t}=-\frac{v^2}{(2013)^2},\ \ \frac t2=2v\Rightarrow v^3=-\frac{2012(2013)^2}{8},\ t=4v.$$

Thus, we know that there is only one such set $(s,t,u,v)$. Hence, the answer is $1$.

mathlove
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