# Chord of a Parabola

## Intersection of a Straight Line with a Parabola

The combined equation of straight line y = mx + c and parabola

y2 = 4ax gives us the co-ordinates of point(s) of their intersection. The combined equation m2x2 + 2x (mc – 2a) + c2 = 0 will give those roots. The straight line therefore meets the parabola at two points.
Points of Intersection of a straight line with the parabola y2 = 4ax
Points of intersection of y2 = 4ax and y = mx + c are given by (mx+c)2=4ax
i.e. m2x2 + 2x(mc – 2a) + c2 = 0. …… (i)
Since (i) is a quadratic equation, the straight line meets the parabola in two points, real, coincident, or imaginary. The roots of (i) are real or imaginary according as {2(mx – 2a)}2 – 4m2c2 is positive or negative, i.e. according as – amc + a2 is positive or negative, i.e. according as mc is less than or greater than a, (taking a as positive).

Note:
When m is very small, one of the roots of equation (i) is very large; when m is equal to zero, this root is infinitely large. Hence every straight line parallel to the axis of the parabola meets the curve in one point at a finite distance and in another point at an infinite distance from the vertex. It means that a line parallel to the axis of theparabola meets the parabola only in one point.

## Length of the chord

As in the preceding article, the abscissae of the points common to the straight line y = mx + c and the parabola y2 = 4ax are given by the equation m2x2 + (2mx – 4a) x + c2 = 0.

Hence, the required length of chord

llustration:
Find the Length of the chord intercepted by the parabola y2 = 4ax from the line y = mx + c. Also find its mid-point. Solution:
Simply by applying the formula o length of the joining (x1, y1) and (x2, y2) we get,
Length of the chord = √((x1-x2 )2+(y1-y2 )2 )
= √((x1-x2 )2+m2 (x1-x2 )2 )
= |x1 – x2| √(1+m2 ) = 4 √(a(a-mc) ) √(1+m2 )
[ ? x1+x2=(-2(m-2a) )/m2 and x1 x2=c2/m2 ]
The midpoint of the chord is ((2a-mc)/m2 ,2a/m)

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