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    Bit stuck on this as appear to have got a really odd value. I assume equation is some kind of ellipse so perhaps weird value is correct?Name:  C1827D1F-36CF-4E47-ACD3-3C4E683B9A28.jpg.jpeg
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    (Original post by Canary84)
    Bit stuck on this as appear to have got a really odd value. I assume equation is some kind of ellipse so perhaps weird value is correct?
    Two things:

    What was the actual question?

    The quality of your images is quite poor, and any potential errors would be difficult to spot.
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    As above, but I think you've put on your second line

     \frac{dy}{dx} = ... = 0.

    But the original equation has no y=... part so the \frac{dy}{dx} part shouldn't be there.
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    Here is the question (number 1) Name:  A49CE44D-8DAB-442D-A230-24ABBDE6E7BF.jpg.jpeg
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    Attempt at re- doing working ( apologies my phone camera is a bit dodgy)
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    (Original post by Canary84)
    Here is the question (number 1)
    Attempt at re- doing working ( apologies my phone camera is a bit dodgy)
    You worked out (x+4y)\frac{dy}{dx}=-(2x+y)............(1)

    But then rearranged it incorrectly to get dy/dx. You have the fraction upside down.

    Should be \frac{dy}{dx}=-\frac{2x+y}{x+4y}

    There is actually no need to perform the division.

    You're looking for when the normal has gradient 4, i.e the tangent has gradient -1/4.

    So, just sub that into (1) for dy/dx. You should get x=0, and then sub into your original equation to get the possible y values.
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    (Original post by ghostwalker)
    You worked out (x+4y)\frac{dy}{dx}=-(2x+y)............(1)

    But then rearranged it incorrectly to get dy/dx. You have the fraction upside down.

    Should be \frac{dy}{dx}=-\frac{2x+y}{x+4y}

    There is actually no need to perform the division.

    You're looking for when the normal has gradient 4, i.e the tangent has gradient -1/4.

    So, just sub that into (1) for dy/dx. You should get x=0, and then sub into your original equation to get the possible y values.
    Thanks for the help!
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