slope of the chord of y^2=8x that gets bisected at (8,2) is - Maths - Conic Sections

Question

slope of the chord of y^2=8x that gets bisected at (8,2) is

Sanjay Kumar Manjhi · Accepted Answer

Dear Student,Equation of parabola: y2=8x=y2=4×2x   (It is in the form of 4ax, where a=2)Now suppose the chord cuts the parabola at P(2p2, 4p) and Q(2q2, 4q)Here P and Q are parametric coordinate
Now slope of chord in parametric form;4p-4q2p2-2q2=4(p-q)2(p2-q2)=4(p-q)2(p-q)(p+q)Slope=2(p+q)Now, the chord is bisected at (8, 2) So, 2p2+2q22=8=p2+q2=8   
(i)similarly,4p+4q2=2=p+q=1=p=1-q  
(ii)Now put value of p in equation (i), we get;p2+q2=8=(1-q)2+q2=8=1+q2-2q+q2=8=2q2-2q-7=0Solving above quadratic equation, we get;q=1±152Now,If q=1+152then p=1-(1+152)=p=1-152Now,Slope=2(p+q)=2(1-152)+(1+152)=41-15+1+15Therefore,Slope=2And P(8-15, 2-215)  Q=(8+15, 2+215)and if q=1-152then p=1-(1-152)=p=1+152Now,Slope=2(p+q)=2(1+152)+(1-152)=41+15+1-15Therefore, Slope=2And P(8+15, 2+215) Q=(8-15, 2-215)Regards