$y=(-3/2)x$ 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
[Math] how to calculate the angle between the tangents of the curve
analytic geometrycalculus
Best Answer
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 $\dfrac {15}{4}$...