Determine the eqn of both lines that are tangent to the graph of f(x)=x^2 and pass through the point (1,-3)
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Determine the eqn of both lines that are tangent to the graph of f(x)=x^2 and pass through the point (1,-3)

[From: ] [author: ] [Date: 13-02-28] [Hit: ]
-3),Since the tangent line also passes through (u, u^2),If u = -1,If u = 3,y = 6x - 9-A line can be written in two ways: parametric or cartesian.......
Please will be greatly appreciated

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Suppose the tangent touches f(x) = x^2 at the point (u, u^2).
Since f'(x) = 2x, then f'(u) = 2u, so the slope of the tangent line is 2u.
So the tangent line is y = 2ux + b, which passes through (1, -3), so we have:
-3 = 2u*1 + b
b = -2u - 3
So the tangent line is y = 2ux - 2u - 3
Since the tangent line also passes through (u, u^2), we have:
u^2 = 2u*u - 2u - 3
u^2 = 2u^2 - 2u - 3
2u^2 - 2u - 3 - u^2 = 0
u^2 - 2u - 3 = 0
u^2 - 3u + u - 3 = 0
u(u - 3) + (u - 3) = 0
(u + 1)(u - 3) = 0
u + 1 = 0 or u - 3 = 0
u = -1 or u = 3

If u = -1, the tangent line is:
y = 2*(-1)*x - 2(-1) - 3
y = -2x + 2 - 3
y = -2x - 1

If u = 3, the tangent line is:
y = 2*3*x - 2*3 - 3
y = 6x - 6 - 3
y = 6x - 9

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A line can be written in two ways: parametric or cartesian.

In the parametric version, you need a point (which your have, P0=(1,-3) ) and a vector, which should be tangent to the function f(x) = x^2
The tangent vector to a function is given by its derivative, here: V0 = (1,2x), where x varies. The connection point of the tangent with the function is still P1= ( x,f(x) ).

So the parametric equation writes:
x = V0x*t + P0x = t+1
y = V0y*t + P0y = 2xt-3

The cartesian equation then is obtained by removing t in the above, taking it from the first equation (i.e. t = x-1 ) and replacing in the second:
y = 2x(x-1)-3

Since this line should also go through the point P1, it comes: 2x(x-1)-3 = x^2, or:
2x^2-2x-3-x^2=0
x^2-2x-3=0, (the start of which looks like a^2-2ab+b^2 ...which is (a-b)^2 )
(x-1)^2 -1 -3 = 0
(x-1)^2 -4 = 0 (which looks like a^2-b^2 = (a-b)(a+b) )
(x-1-2)(x-1+2) = 0

which has 2 solutions: x = 3 and x = -1

Coming back to the parametric equation above we have the 2 lines as follows:
x = t+1
y = 2*3*t-3 = 6t-3
and
x = t+1
y = 2*(-1)*t-3 = -2t-3

both of which individually transform into (remove the t in each)
y = 6(x-1)-3, or y = 6x-9
and
y = -2(x-1)-3, or y = -2x-1

You can verify that the point (1,-3) is on these lines and that the lines intersect the function at P1, i.e. (3,9) for line one and (-1,1) for line two.
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