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Thread: find specific points of a tangent line on a curve.

  1. #1
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    find specific points of a tangent line on a curve.

    Q: Find the coordinates of all points on the graph of $\displaystyle y=1-x^2$ at which the tangent line passes through the point $\displaystyle (2,0)$

    My solution so far:

    What I understood is that there is a point on outside the parabola/ curve $\displaystyle (2,0)$ and another point is on the curve which is unknown and need to figure out $\displaystyle (x, y)$ and they are connected through a tangent line.

    Now first let me figure out the derivative/slope of the tangent line over that curve.

    $\displaystyle y=1-x^2$

    $\displaystyle f(x)=1-x^2$ { is y always equals f(x)? maybe this is only when y is function of x right?}

    $\displaystyle f'(x)=-2x$

    So now we know the slope of the tangent line
    we know 2 co-ordinates of one end $\displaystyle (2,0)$ and need to figure out the other end $\displaystyle (x,y)$.

    and we can use the slope formula here

    $\displaystyle \frac{0-y}{2-x}=-2x$

    $\displaystyle -y=-2x(2-x)$

    $\displaystyle -y=-4x+2x^2$

    $\displaystyle -(1-x^2)=-4x+2x^2$ [as $\displaystyle 1-x^2=y$ from the original equation]

    $\displaystyle 1-x^2=4x-2x^2$

    $\displaystyle 0=4x-2x^2+x^2-1$

    $\displaystyle 0=4x-x^2-1$

    $\displaystyle 0=-x^2+4x-1$

    No easy way to factor this so I used the formula

    $\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

    So that makes it

    $\displaystyle x=\frac{-4\pm\sqrt{4^2-4(-1)(-1)}}{2(-1)}$

    $\displaystyle x=\frac{-4\pm\sqrt{16-4}}{-2}$

    $\displaystyle x=2\pm\sqrt{12}$

    $\displaystyle x=2\pm2\sqrt{3}$

    so basically we got 2 xs

    $\displaystyle x=2+2\sqrt{3}$ and $\displaystyle x=2-2\sqrt{3}$


    where did I do wrong..

    why the answer says $\displaystyle 2\pm\sqrt{3}$ not sure if that is x or y.. looks close to x...
    Last edited by ameerulislam; Apr 26th 2013 at 05:22 AM.
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  2. #2
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    Re: find specific points of a tangent line on a curve.

    Quote Originally Posted by ameerulislam View Post
    Q: Find the coordinates of all points on the graph of $\displaystyle y=1-x^2$ at which the tangent line passes through the point $\displaystyle (2,0)$

    My solution so far:

    What I understood is that there is a point on outside the parabola/ curve $\displaystyle (2,0)$ and another point is on the curve which is unknown and need to figure out $\displaystyle (x, y)$ and they are connected through a tangent line.

    Now first let me figure out the derivative/slope of the tangent line over that curve.

    $\displaystyle y=1-x^2$

    $\displaystyle f(x)=1-x^2$ { is y always equals f(x)? maybe this is only when y is function of x right?}

    $\displaystyle f'(x)=-2x$

    So now we know the slope of the tangent line
    we know 2 co-ordinates of one end $\displaystyle (2,0)$ and need to figure out the other end $\displaystyle (x,y)$.

    and we can use the slope formula here

    $\displaystyle \frac{0-y}{2-x}=-2x$

    $\displaystyle x=2\pm2\sqrt{3}$ CORRECT
    BUT $\displaystyle x=2\pm2\sqrt{3}\ne \pm 4\sqrt{3}$

    $\displaystyle (2\pm2\sqrt{3})^2=4\pm4\sqrt{3}+12$
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  3. #3
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    Re: find specific points of a tangent line on a curve.

    Quote Originally Posted by Plato View Post
    BUT $\displaystyle x=2\pm2\sqrt{3}\ne \pm 4\sqrt{3}$

    $\displaystyle (2\pm2\sqrt{3})^2=4\pm4\sqrt{3}+12$
    yeah yeah i figured that our immediately and edited., but why the bok is saying $\displaystyle x=2\pm\sqrt{3}$ 2 is missing. so they are wrong and I didn't make any other mistakes?
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  4. #4
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    Re: find specific points of a tangent line on a curve.

    ok I think I found it..

    that basic mistake was done up there as well.
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