Answer to Question #105714 in Geometry for Akanksha

Question

Normals at any point P of ellipsoid x2/9+y2/4+z2=1 meet planes in Q1,Q2,Q3 resp. Show PQ1:PQ2:PQ3=9:4:1

Answer 1

Equation of normal to the ellipse $ax^2+by^2+cz^2=1$ is $\frac{x-\alpha}{a\alpha}=\frac{y-\beta}{b\beta}=\frac{z-\gamma}{c\gamma}$

For the given elllipse $a=\frac{1}{9} ; b=\frac{1}{4} ; c=1$

$\frac{x-x_o}{x_o/9}=\frac{y-y_o}{y_o/4}=\frac{z-z_o}{z_o}$

When $x=0;y=\frac{-5y_o}{4};z=-8z_o$

$Q_1=(0,\frac{-5y_o}{4},-8z_o)$

When $z=0;x=8x_o/9;y=3y_o/4$

$Q_3=(8x_o/9,3y_o/4,0)$

When $y=0; x=\frac{5x_o}{9};z=-3z_o$

$Q_2=(\frac{5x_o}{9},0,-3z_o)$

$P=(x_o,y_o,z_o)$

$PQ_1=\sqrt{(x_o-0)^2+(y_o+\frac{5y_o}{4})^2+(z_o+8z_o)^2}$ $PQ_1=\sqrt{(x_o)^2+(\frac{9y_o}{4})^2+(9z_o)^2}$

$PQ_1=\sqrt{16x_o^2+81y_o^2+36^2z_o^2}/4$

$PQ_3=\sqrt{(x_o-8x_o/9)^2+(y_o-3y_o/4)^2+(z_o-0)^2}$ $PQ_3=\sqrt{16x_o^2+81y_o^2+36^2z_o^2}/36$

$PQ_2=\sqrt{(x_o-\frac{5x_o}{9})^2+(y_o-0)^2+(z_o+3z_o)^2}$ $PQ_2=\sqrt{(\frac{4x_o}{9})^2+(y_o)^2+(4z_o)^2}$

$PQ_2=\sqrt{16x_o^2+81y_o^2+36^2z_o^2}/9$

$PQ_1:PQ_2:PQ_3=1/4:1/9:1/36$

$PQ_1:PQ_2:PQ_3=9:4:1$