Question

equation of the locus of the midpoint of the focal chord of the parabola y^2=64x

Answer 1

equation one that is X is equal to 64 that equation to Y is equal to 2 then you have to divide a equation 1 to 4 x + Y is equal to 64 / 2 then you will get finally answer 32 .therefore the midpoint of the focal chord of the parabola is 32

Answer 2

Answer:

y² = 32(x-16).

Step-by-step explanation:

Focus, S of the given parabola is (16,0).

Equaton of focal chord, that is a chord of a parabola which is passing through its fous is

   y = m(x-16)-------(*), if we assume 'm' to be its slope.

For, equation(*) to be the chord of a parabola it has to intersect the parabola in 2 distinct points, let those be A (x₁, y₁) and B(x₂, y₂).

Solving y²=64x and y = m(x-16) we get

  m²(x-16)²=64x,

 m²x² -32x(m²+2) + 256m² = 0, whose roots are x1, x2.

So, Abcissa of the midpoint, X = (x₁ +x₂)/2 = 16(m²+2)/m²-----(1)

Ordinate of the midpoint, Y = m(x₁-16 +x₂-16)/2,

  Y=m(X - 16)-----(2),

On eliminating m from eq(1) and (2), we get

y² = 32(x-16), which is the required locus of the midpoint of the focal chord of the given parabola.