Find the locus of the middle points of the chords of the hyperbola $3x^2-2y^2+4x-6y=0$ parallel to the line $y=2x$ .
Equation of the chords of the hyperbola bisected at $(h,k)$ is given by $S_1=T$ where $S_1=3xx_1-2yy_1+2(x+x_1)-3(y+y_1)=3x_1^2-2y_1^2+4x_1-6y_1$
where I have chosen $(x_1,y_1)$ to be any point of the line $y=2x$
But the answer is not coming.Is there any fault in my answer.
Please help.
Best Answer
Let $y=2x+k$ be an equation of the chord and $(p,q)$ be a point on our locus.
Thus, for the $x$-coordinate of an intersect point we obtain: $$3x^2-2(2x+k)^2+4x-6(2x+k)=0$$ or $$5x^2+8(1+k)x+2k^2+6k=0,$$ which gives $$x_1+x_2=\frac{-8(1+k)}{5}$$ and $$p=\frac{-4(1+k)}{5}$$ and $$q=2p+k=\frac{-8(1+k)}{5}+k=\frac{-3k-8}{5}.$$ Id est, $$k=\frac{5p+4}{-4}=\frac{5q+8}{-3},$$ which gives $$q=\frac{3}{4}p-1$$ and we got an equation of the locus: $$y=\frac{3}{4}x-1.$$