# Math Help - Equation Of A Tangent Q

1. ## Equation Of A Tangent Q

ok so here's the Q i am stuck on

1) Find the equations of all tangents to the graph of f(x) = x^2 - 4x + 25 that passes through the origin (0,0)

so from my understanding, the tangent line has to pass through the origin

y = mx + b

sub in (0,0) gives me b = 0 (makes sense since line isnt shifted and only the slope is something diff)

so then i differentiate the f(x)
and get f'(x) = 2x-4, which is the slope

then i sub that back into my line equation

y = (2x-4)x + b

y = 2x^2 -4x +b

0 = 2x^2 -4x

solving for x i get x = 2, x = 0

up until here i am wondering if i did this correctly

and if i did, how do i now write the equation of the tangents down?

y = 2x + b AND y = b

??

2. Hello, kamkal!

You're mixing your x's . . .

1) Find the equations of all tangents to the graph of: . $f(x) \:= \:x^2 - 4x + 25$
that passes through the origin (0,0).

Let the point $A(p,q)$ be on the parabola.
. . Note that: . $q \:=\:p^2 - 4p + 25$ .[1]

A tangent to the curve has slope: . $f'(x) \:=\:2x-4$

At point $A(p,q)$, the slope is: . $m \:=\:2p - 4$

The equation of the line through $A(p,q)$ with slope $2p-4$ is:

. . $y - q \;=\;(2p-4)(x - p) \quad\Rightarrow\quad y \;=\;(2p-4)x - 2p^2 + 4p + q$

Substitute [1]: . $y \;=\;(2p-4)x - 2p^2 + 4p + (p^2 - 4p + 25)$

. . which simplifies to: . $y \;=\;(2p-4)x - p^2 + 25$ .[2]

Since $(0,0)$ is on this tangent, we have:
. . $0 \;=\;(2p-4)\!\cdot\!0 - p^2 + 25 \quad\Rightarrow\quad p^2 \:=\:25 \quad\Rightarrow\quad p \:=\:\pm5$

Substitute into [2]:

$\begin{array}{cccccccc}p = 5\!: & y &=& (10-4)x - 5^2 + 25 & \Rightarrow & {\color{blue}y \;=\; 6x} \\ \\[-3mm]
p = \text{-}5\!: & y &=& (\text{-}10-4)x - (\text{-}5)^2+25 & \Rightarrow & {\color{blue}y \;=\; \text{-}14x} \end{array}$

3. Originally Posted by kamkal
ok so here's the Q i am stuck on

1) Find the equations of all tangents to the graph of f(x) = x^2 - 4x + 25 that passes through the origin (0,0)

so from my understanding, the tangent line has to pass through the origin

y = mx + b

sub in (0,0) gives me b = 0 (makes sense since line isnt shifted and only the slope is something diff)

...
Here is another way to solve the question:

The tangent has the equation

$t: y = m\cdot x$

Now calculate the points of intersection between the parabola and t. You are looking for those values of m where the 2 graphs have exactly one point in common:

$x^2-4x+25 = mx~\implies~x^2-(4+m)x+25=0$

$x = \dfrac{-\frac12(4+m)\pm\sqrt{\frac14(4+m)^2-25}}{2}$

If there exists only one point of intersection then the discriminante (=the term under the square-root sign) has to be zero:

$\frac14(4+m)^2-25=0~\implies~\frac14m^2+2m-21=0~\implies m=-14~\vee~m=6$

That means the tangents have the equations:

$t_1: y=-14x$

$t_2: y = 6x$