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    Okay so I have a parabola with equation y^2=x and I need to find the tangent to the parabola at the point (16,-4) I tried setting y^2=(mx+c)^2=0 and then trying to get m using that b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

    Any advice?

    I will check back in \sim 1hr.
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    (Original post by poorform)
    Okay so I have a parabola with equation y^2=x and I need to find the tangent to the parabola at the point (16,-4) I tried setting y^2=(mx+c)^2=0 and then trying to get m using that b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

    Any advice?

    I will check back in \sim 1hr.
    Why have you set that expression equal to 0?

    Just substitute y = mx + c into the original and solve.

    Alternatively, work out the gradient of the parabola at the given point and use that to work out the equation of the tangent
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    (Original post by poorform)
    Okay so I have a parabola with equation y^2=x and I need to find the tangent to the parabola at the point (16,-4) I tried setting y^2=(mx+c)^2=0 and then trying to get m using that b^2-4ac=0 since it's a tangent but that didn't work and I don't really know what to do now.

    Any advice?

    I will check back in \sim 1hr.
    My initial thoughts would be to differentiate y^2=x and rearrange to get \frac{dy}{dx}=.... Plug in numbers to get gradient of the tangent at that point i.e. m. Then you can use y-y_1 = m(x-x_1), where (x_1, y_1) is the point they've given you.
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    (Original post by davros)
    Why have you set that expression equal to 0?

    Just substitute y = mx + c into the original and solve.

    Alternatively, work out the gradient of the parabola at the given point and use that to work out the equation of the tangent
    I'm fine using the second method but I'd prefer not to use any calculus at this point. I still can't figure out what to do could you look through what I've done and help.

    \displaystyle y^2=x

    \displaystyle y=mx+c

    \displaystyle y^2=(mx+c)^2=m^2x^2+2mxc+c^2-x the line is a tangent \displaystyle \implies discriminant must be equal to zero and so \displaystyle (2xc)^2-4x^2(c^2-x)=0 \iff 4x^2c^2-4x^2c^2-4x^3=0 \iff -4x^3=0.

    Have I done something wrong I'm not too sure I understand exactly what it going on here so if you could help me I would appreciate it. Thanks.
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    (Original post by poorform)
    I'm fine using the second method but I'd prefer not to use any calculus at this point. I still can't figure out what to do could you look through what I've done and help.

    \displaystyle y^2=x

    \displaystyle y=mx+c

    \displaystyle y^2=(mx+c)^2=m^2x^2+2mxc+c^2-x the line is a tangent \displaystyle \implies discriminant must be equal to zero and so \displaystyle (2xc)^2-4x^2(c^2-x)=0 \iff 4x^2c^2-4x^2c^2-4x^3=0 \iff -4x^3=0.

    Have I done something wrong I'm not too sure I understand exactly what it going on here so if you could help me I would appreciate it. Thanks.
    Anyone?
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    (Original post by poorform)
    Anyone?
    Nevermind got it.
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    (Original post by poorform)
    Anyone?
    x shouldn't be part of the discriminant since you have a quadratic in the variable x!

    (Original post by poorform)
    Nevermind got it.
    Good stuff
 
 
 
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