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    Here is a question I have been given in my exams:

    Name:  ma555 8.png
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    I have an answer but as I have no resources to check it against, I am posting my answers on here to try and get some extra help. Here is my answer:

    a) f(x,y,z)= \frac{x}{a}+\frac{y}{b}+ \frac{z}{c}, (\frac{x}{a})^2+(\frac{y}{b})^2+  (\frac{z}{c})^2=1

    So  g(x,y,z)=(\frac{x}{a})^2+(\frac{  y}{b})^2+(\frac{z}{c})^2

     F_x={\lambda}g_x, \Longrightarrow \frac{1}{a}=\lambda(\frac{2x}{a^  2}), \lambda=\frac{a}{2x} Call this equation (1).

     F_y={\lambda}g_y, \Longrightarrow \frac{1}{b}=\lambda(\frac{2y}{b^  2}), \lambda=\frac{b}{2y} Call this equation (2).

    (1)=(2) \Longrightarrow \frac{a}{2x}=\frac{b}{2y} \Longrightarrow x=\frac{ay}{b}

     F_z={\lambda}g_z, \Longrightarrow \frac{1}{c}=\lambda(\frac{2z}{c^  2}), \lambda=\frac{c}{2z} Call this (3).

    (3)=(2)  \Longrightarrow z=\frac{cy}{b}

    So: (\frac{a\frac{y}{b}}{a})^2+(y/b)^2+(\frac{c\frac{y}{b}}{c})^2=  1

    This gives:\frac{3y^2}{b^2}=1, b= \pm \sqrt{3}y \Longrightarrow x=\frac{\pm a}{\sqrt{3}}, z=\frac{\pm c}{\sqrt{3}}

    I think this is the correct method, I do not however know how to tackle b. Thanks for any help!
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    (Original post by Happy2Guys1Hammer)
    Here is a question I have been given in my exams:

    Name:  ma555 8.png
Views: 47
Size:  42.2 KB,

    I have an answer but as I have no resources to check it against, I am posting my answers on here to try and get some extra help. Here is my answer:

    a) f(x,y,z)= \frac{x}{a}+\frac{y}{b}+ \frac{z}{c}, (\frac{x}{a})^2+(\frac{y}{b})^2+  (\frac{z}{c})^2=1

    So  g(x,y,z)=(\frac{x}{a})^2+(\frac{  y}{b})^2+(\frac{z}{c})^2

     F_x={\lambda}g_x, \Longrightarrow \frac{1}{a}=\lambda(\frac{2x}{a^  2}), \lambda=\frac{a}{2x} Call this equation (1).

     F_y={\lambda}g_y, \Longrightarrow \frac{1}{b}=\lambda(\frac{2y}{b^  2}), \lambda=\frac{b}{2y} Call this equation (2).

    (1)=(2) \Longrightarrow \frac{a}{2x}=\frac{b}{2y} \Longrightarrow x=\frac{ay}{b}

     F_z={\lambda}g_z, \Longrightarrow \frac{1}{c}=\lambda(\frac{2z}{c^  2}), \lambda=\frac{c}{2z} Call this (3).

    (3)=(2)  \Longrightarrow z=\frac{cy}{b}

    So: (\frac{a\frac{y}{b}}{a})^2+(y/b)^2+(\frac{c\frac{y}{b}}{c})^2=  1

    This gives:\frac{3y^2}{b^2}=1, b= \pm \sqrt{3}y \Longrightarrow x=\frac{\pm a}{\sqrt{3}}, z=\frac{\pm c}{\sqrt{3}}

    I think this is the correct method, I do not however know how to tackle b. Thanks for any help!
    Form three triangles by joining P to the three vertices. What are the areas of these triangles? Add them up to get the given expression.
 
 
 
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