Equation of tangent lines?

Question

Some lines which are tangent to the parabola y=x^2 also pass through the point (2,3). Find all of these lines.

Can someone guide me on how to do this?

Answer 1

y' = 2x
Two points: (x, x^2) and (2,3)
(x^2-3)/(x-2) = 2x
x^2-3 = 2x^2-4x
x^2-4x+3 = (x-1)(x-3) = 0
x = 1, 3

For x = 1, y' = 2, (x, x^2) = (1, 1)
y = 2(x-1)+1 = 2x - 1

For x = 3, y' = 6,
Since (2,3) is a point on the line, we have
y = 6(x-2)+3 = 6x - 9

Answer 2

There are two such tangent lines. They will have the same slope as the curve at the point of tangency. First find the slope of the curve. To do that take the derivative.

y = x²
dy/dx = 2x

Find the equation of the line thru the point (2,3) and tangent to the curve y = x² at the point (x, x²).

y - 3 = 2x(x - 2)
x² - 3 = 2x(x - 2) = 2x² - 4x
0 = x² - 4x + 3
(x - 1)(x - 3) = 0
x = 1, 3

The points of tangency are (1, 1²) and (3, 3²). That is they are (1, 1) and (3, 9).

Now write the equation of the tangent lines thru (2, 3).

y - 3 = 2*1(x - 2)
y - 3 = 2x - 4
y = 2x - 1
_______

y - 3 = 2*3(x - 2)
y - 3 = 6x - 12
y = 6x - 9

Answer 3

Take y-3=m(x-2) and intercept with y=x^2
x^2-3 = mx-2m so
x^2-mx+2m-3=0 gives you intercepts
But there must be only one intercept
so
m^2-4(2m-3)=0 ( discriminant)
m^2-8m+1 2= 0 so m=(( 8+- sqrt( 64-48))/2
so m= 6 and m = 2
so y-3=6(x-2) and y-3=2(x-2)