Find the equation of the straight line(s) which pass through the point (1, −2) and is (are) tangent to the parabola with equation y = x^2

Question

Find the equation of the straight line(s) which pass through the point (1, −2) and is (are) tangent to the parabola with equation y = x^2

Answer 1

line through (1,-2) tangent to y=x^2

the line is y=mx +b

-2=m + b, m= y' = 2x = slope of the parabola

the tangent line goes through the point (x,y) = (x,x^2) and (1,-2)

which have slope (x^2+2)/x-1)=2x

2x^2 -2x = x^2+2

x^2 -2x -2 = 0

x^2-2x+1 = 2+1 =3

(x-1)^2 = 3

x-1 = +/-sqr3

x = 1+/-sqr3

y =x^2= 4+2sqr3 or4-2sqr3

tangency points are (1+sqr3, 4+2sqr3), (1-sqr3, 4-2sqr3)

with slope = 2x = 2+2sqr3 or 2-2sqr3

tangent lines are in point slope form

y-4-2sqr3 = (2-2sqr3)(x-1+sqr3)

downward slopeing

and

y -4+2sqr3 = (2+2sqr3)(x-1-sqr3)

upward sloping