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    Stuck of part D, even after looking solutionbank, the rest is fine.
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    An explanation will be good!
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    (Original post by P____P)
    Stuck of part D, even after looking solutionbank, the rest is fine.
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    An explanation will be good!
    From the tangent equation, get the gradient of the tangent. Then do -1/that gradient to get the gradient of the perpendicular (which we'll call m). Since the perpendicular passes through the origin, its equation is simply y = mx. Then substitute this into the tangent equation to get the point of intersection, which will be the foot of the perpendicular.
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    (Original post by HapaxOromenon2)
    From the tangent equation, get the gradient of the tangent. Then do -1/that gradient to get the gradient of the perpendicular (which we'll call m). Since the perpendicular passes through the origin, its equation is simply y = mx. Then substitute this into the tangent equation to get the point of intersection, which will be the foot of the perpendicular.

    Right. Wont the intersection just be the point P?
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    (Original post by HapaxOromenon2)
    From the tangent equation, get the gradient of the tangent. Then do -1/that gradient to get the gradient of the perpendicular (which we'll call m). Since the perpendicular passes through the origin, its equation is simply y = mx. Then substitute this into the tangent equation to get the point of intersection, which will be the foot of the perpendicular.
    (Original post by P____P)
    Right. Wont the intersection just be the point P?
    Never mind. It isnt!
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    Following on from HapaxOromenon2's post. Having found the coordinates for the foot of the perpendicular, work out the lefthand side of the equation you're trying to show, and work out the right hand side, and show they're equal.

    Coordinates for the foot are:
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    \displaystyle x=\frac{12\cos\theta}{4\cos^2  \theta+9\sin^2\theta}

    \displaystyle y=\frac{18\sin\theta}{4\cos^2  \theta+9\sin^2\theta}

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    (Original post by P____P)
    Right. Wont the intersection just be the point P?
    only if the ellipse is circular
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    (Original post by ghostwalker)
    Following on from HapaxOromenon2's post. Having found the coordinates for the foot of the perpendicular, work out the lefthand side of the equation you're trying to show, and work out the right hand side, and show they're equal.

    Coordinates for the foot are:
    Spoiler:
    Show


    \displaystyle x=\frac{12\cos\theta}{4\cos^2  \theta+9\sin^2\theta}

    \displaystyle y=\frac{18\sin\theta}{4\cos^2  \theta+9\sin^2\theta}

    Thanks for your help. Usually you use trig identities to manipulate the x's and the y's to match what is required, I didn't know you could just substitute basically.

    (Original post by the bear)
    only if the ellipse is circular
    hmm, thanks.

    (Original post by HapaxOromenon2)
    From the tangent equation, get the gradient of the tangent. Then do -1/that gradient to get the gradient of the perpendicular (which we'll call m). Since the perpendicular passes through the origin, its equation is simply y = mx. Then substitute this into the tangent equation to get the point of intersection, which will be the foot of the perpendicular.
    thanks.
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    Made a graph just for fun!
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    varying the parameter a gives the locus that needs to be proven
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    (Original post by P____P)
    Made a graph just for fun!
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    varying the parameter a gives the locus that needs to be proven
    Very nice. Well done.
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    (Original post by Zacken)
    Very nice. Well done.
 
 
 
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