# Thread: equations of the 2 tangent lines

1. ## equations of the 2 tangent lines

find equations of 2 tangent lines passing through (2,5)

f(x)= 4x-x^2

points of tangent line are (3,3) and (1,3)

2. Originally Posted by phalange
find equations of 2 tangent lines passing through (2,5)

f(x)= 4x-x^2

points of tangent line are (3,3) and (1,3)
hi phalange,

the derivative of the function gives the slope of the tangent for some x.

hence 4-2x=slope.

The points are (x,y)=(x,f(x)).

The line equations are

$y-y_1=m(x-x_1)$ for all x

where m is $4-2x_1$ for $x_1=3\ and\ 1$

3. Since you are given the points of tangency, that is particularly easy. If you were only given that the tangent lines went through (2, 5), it would be harder. Suppose the point of tangency is $(x_0, 4x_0- 2x_0^2)$. The derivative at $x_0$ is 4- 4x_0 so the equation of the tangent line there is $y= (4- 4x_0)(x- x_0)+ 4x_0- x_0^2= 4x- 2x_0x+ x_0^2$. At x= 2, that is $y= 8- 4x_0+ x_0^2= 5$ or $x_0^2- 4x_0+ 3= (x_0-3)(x_0- 1)= 0$. That gives x= 1 and x= 3 as the points of tangency.

Here is Fermat's pre-calculus (literally!) method:

Any non-vertical line through (2, 5) can be written as y= m(x-2)+ 5. In order that such a line be tangent to $y= 4x- x^2$, they must, of course, touch: [tex]y= 4x- x^2= m(x-2)+ 5[tex] for some x= a. That is the same as saying $x^2+ (m- 4)x- 2m+ 5= 0$ has x=a as solution and so has (x-a) as a factor.

To be tangent a must be a double root and x-a must be a double root. That means that we must have
$x^2+ (m- 4)x+ (5- 2m)= (x- a)^2= x^2- 2ax+ a^2$ for all x.

Set coefficients equal and solve for m.

m- 4= -2a and $5- 2m= a^2$. Since m= 4- 2a, 2m= 8- 4a and $5- 2m= 5- 8+ 4a= -3+ 4a= a^2$ or $z^2- 4a+ 3= 0$ as before.

4. That was beautifully portrayed, HallsofIvy!

thank you,
much appreciated!