Question

Prove the chord 4x+2y+1=0 of parabola y^2=8x is bisected at the point (2,-3)

Answer 1

Answer:

Step-by-step explanation:

given that,

y^2 = 8x

x = y^2/8

Now,

4x + 2y + 1 = 0

putting the value of x

4(y^2/8) + 2y + 1 = 0

y^2/2 + 2y +1 = 0

y^2 + 4y + 2 = 0

solve the quadratic equation and you will get the answer!!

Hope this helped you.....