# how to calculate the angle between the tangents of the curve?

$y=(-3/2)x$ and $y=(-2/5)x$ intersect the curve $$3x^2+4xy+5y^2-4=0$$ at points $P$ and $Q$ .find the angle between tangents drawn to curve at $P$ and $Q$ .I know a very long method of finding intersection points then differentiating to find the slope of two tangents and then finding the angle between them .Is there any shorter and elegant method for questions like these, like using some property of curve . Thanks in advance

• If the two lines were themselves tangent, you could calculate immediately from the given slopes. If they aren't tangent, then you have to find $P,Q$ as stated and do everything the difficult way... – abiessu May 17 '14 at 14:19
• Sadly they are'nt , – avz2611 May 17 '14 at 14:31

You can find it without finding $P,Q$.

By implicit derivative we have, $$6x+4y+4xy'+10y=0$$ $$y'=\frac {-14y}{4x}-\frac {3}{2}$$ if you put $-\dfrac{3x}{2}$ into $y$ above equation you will find the slope of line passing through the $P$ which is $\dfrac {15}{4}$...

• Very nice, indeed. – Claude Leibovici May 17 '14 at 14:52

Hint

Finding the coordinates of points $P$ and $Q$ is not difficult at all since the problem reduces to the solution of two quadratic equations. By the way, the intersection of $3x^2+4xy+5y^2-4=0$ and $y=k x$ are given by $$x=\pm \frac{2}{\sqrt{5 k^2+4 k+3}},y=\pm \frac{2k}{\sqrt{5 k^2+4 k+3}}$$

Then, use implicit differentiation to get the slopes at these points and finally compute the angle.

I am sure that you can take from here.