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I have such an exercise:

Find the measure of angle formed by the tangent lines drawn through $A(2,-1)$ to the following function: $$f:R\to R, f(x)=x^2$$

My solving was going well till I got stuck at the point when I found that we have two tangent lines drawn through the point I mentioned about($A$). So I 've got this:

$$y=x_0^2+(4+2\sqrt{5})(x-x_0)$$ and $$y=x_1^2+(4-2\sqrt{5})(x-x_1)$$

Respectively $x_0$ represents the points in which the lines above are tangent, this means: $$x_0=2+\sqrt5$$ and $$x_1=2-\sqrt5$$

My question is : how can I find the measure of angle formed by this two tangent lines?

I hope you'll understand the problem and share you ideas. Thank you very much!

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2 Answers 2

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If $m_1$ and $m_2$ are the gradients of the two lines, the angle between them (assuming they are not parallel) is given by $\displaystyle \tan \theta = | \frac{m_1 - m_2}{1+m_1m_2} |$.

You know the x-coordinates of the tangents so finding the gradients is straight-forward.

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Hint: Find a direction vector for each line. Then use the formula $$\cos\angle(\vec u, \vec v)=\frac{|\vec u\cdot \vec v|}{|\vec u|\cdot|\vec v|}$$

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  • $\begingroup$ Thank you but, are there any other possibilities, because I have never worked much with vectors? $\endgroup$
    – wonderingdev
    Apr 20, 2014 at 13:47

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