Question

The midpoint of the chord 2x+y-4=0of the parabola y^2=4x is

Answer 1

locus of parabola y² = 4ax is (at², 2at)

so, point (t², 2t) lies on parabola.

Let point (t², 2t) is the end point of chord.
so, point (t², 2t) satisfies equation of chord 2x + y - 4 = 0
now, 2t² + 2t - 4 = 0
t² + t - 2 = 0
t² + 2t - t - 2 = 0
t(t + 2) - 1(t + 2) = 0
(t - 1)(t + 2) = 0
t = -2, 1

hence, end points are (1², 2 × 1) and {(-2)², 2×-2}
e.g., (1, 2) and (4, -4)
so, the midpoint of chord is {(1 + 4)/2, (2 - 4)/2}
= (5/2, -1)

hence, answer is (5/2, -1)

Answer 2

Solution:

Equation of the chord 2x+y+4=0 y= 4-2x.....(1)

Equation of parabola y² = 4x......(2)  

First we have to find the point of intersection of the chord and parabola. using (1) in (2)

(4-2x)² = 4x  

4(2-x)² =4x  

(2-x)² =x  

x²-4x+4 =x  

x²-5x+4 =0  

(x-1)(x-4)=0  

x=1, 4  

when x=1, y=4-2=2 when x=4, y=4-8= - 4

 Therefore the end points of the chord are(1,2) and (4,-4)  

The midpoint of the chord is

$(\frac{{x_1}+{x_2}}{2},\frac{{y_1}+{y_2}}{2})\\\\(\frac{1+4}{2},\frac{2-4}{2})\\\\(\frac{5}{2},\frac{-2}{2})\\\\(\frac{5}{2},-1)$