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

Question

$y=(-3/2)x$ 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

Answer 1

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 $\dfrac {15}{4}$...

Answer 2

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.