A straight line passes through the point (2, 0) and has gradient m. Find the two values of m for which the line is a tangent to the curve y = x^2 − 4x + 5. For each value of m, find the coordinates of the point where the line touches the curve.
The gradient(n) of the tangent for the curve is given by;
`dy/dx = 2x-4`
The equation of line passing (2,0) is given by;
`y = mx+c`
Since the line passes through (2,0)
`0 = 2xm+c`
`c = -2m`
The equation of the line is;
`y = mx-2m`
If the line is tangent to the curve it satisfies the gradient of tangent.
At the tangent point
`2x^2-8x+8 = x^2-4x+5`
`x^2-4x+3 = 0`
`x=3` and `x=1`
When `x=3` then `m=2xx3-4=2`
When `x=1` then `m=2xx1-4=-2`
When x = 3 then `y = 2^2 − 4xx2 + 5 = 1`
When `x = 1` then `y = 1^2 − 4xx1 + 5 = 2`
So the two values of in which the straight line is a tangent to curve is m=2 and m=-2.
The curve touches the line at points (3,1) and (1,2).