Question

Find the values of m for which the lines y=mx-2 are tangents to the curve with equation y=x^2-4x+2

Answer 1

Let y= mx-2 be the tangent line to y=x^2-4x+2 at x=a.

Then slope, $m=\frac{dy}{dx} at x=a = 2a-4.$

Hence the equation is y=(2a-4)x-2

Let's find y-coordinate at x=a using tangent line and curve.

Using tangent line y at x=a is (2a-4) a -2 $=2a^{2}-4a-2$

Using given curve y-coordinate at x=a is $a^{2}-4a+2$

Let's equate these 2 y-coordinates,

 $2a^{2} -4a-2 = a^{2} -4a+2$

$2a^{2}-a^{2} = 2+2$

$a^{2}=4$

a=2 or -2.

If a=2, $m=2a-4 = 2*2-4=0$

If a=-2,$m= 2(-2)-4 = -8$

Hence m values are 0 and -8.