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Math Help - Equation Of A Tangent Q

  1. #1
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    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

    ??

    thanks in advance
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  2. #2
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    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]<br />
p = \text{-}5\!: & y &=& (\text{-}10-4)x - (\text{-}5)^2+25 & \Rightarrow &  {\color{blue}y \;=\; \text{-}14x} \end{array}

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  3. #3
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    Quote Originally Posted by kamkal View Post
    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

    Use the quadratic formula:

    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
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