A straight line moves so as to meet the straight lines $y=mx, z=c$ and $y=-mx, z=-c$ in A and B and intersects the curve $yz=k^2, x=0$, show that the locus of the middle point of $AB$ is $$(m^2x^2-y^2)mc=k^2y^2$$
Please give me some hints to solve the problem.
The straight line has a point $A(a,ma,c)$ on $y=mx,z=c$, a point $B(b,-mb,-c)$ on $y=-mx,z=-c$ and a point $C(0,y,z)$ on $yz=k^2,x=0$.
These are collinear so looking at the $x$-coordinates we must have $(b-a)C=bA-aB$ and hence $C=(0,\frac{2mab}{b-a},c\frac{a+b}{b-a})$. Since this lies on the curve we have $2mab(a+b)c=k^2(b-a)^2\ (*)$.
The midpoint of $AB$ is $\left(\frac{1}{2}(a+b),\frac{1}{2}m(a-b),0)\right)$. This lies in the plane $z=0$ and the $x,y$ coordinates satisfy $m^2x^2-y^2=abm^2$ and $4y^2=m^2(b-a)^2$ and $2x=(a+b)$. So from $(*)$ we have $$(m^2x^2-y^2)mcx=k^2y^2$$
Note that this is significantly different from what is stated in the question.