Question

the tangent line to the curve y=x^2 at the point (a,a^2) passes through the point (2,1) find all possible values of a.
Thank you so much!!

Answer 1

dy/dx = 2x
at (a,a^2) , dy/dx = 2a

equation of tangent:
y - a^2 = 2a(x-a)
at 2,1)
1 - a^2 = 2a(2-a)
1 - a^2 = 4a - 2a^2
a^2 - 4a + 1 = 0
let's complete the square:
a^2 - 4a + 4 = -1 + 4
(a-2)^2 = 3
a-2 = ± √3
a = 2 ± √3

or

let point of contact be (a,a^2) , let P(2,1) be the outside point.
dy/dx = 2a
so at (a,a^2) slope = 2a
grade 9 way: slope = (a^2-1)/(a-2)

so (a^2 - 1)/(a-2) = 2a
2a^2 - 4a = a^2 - 1
a^2 - 4a + 1 = 0
same as before