Answer to Question #114946 in Analytic Geometry for ANJU JAYACHANDRAN

Question

The normals at any point P of the ellipsoid x^2/9 +y^2+4 +z^2 = 1 meet the coordinate planes in Q1,Q2,Q3, respectively. Show that PQ1 : PQ2 : PQ3 :: 9 : 4 : 1.

Answer 1

The equation of the ellipsoid is $\dfrac{x^2}{9}+\dfrac{y^2}{4}+z^2=1$.

Equation of normal at any point $P=(x_0,y_0,z_0)$ to the ellipsoid $ax^2+by^2+cz^2=1$ is $\frac{x-x_0}{ax_0}=\frac{y-y_0}{by_0}=\frac{z-z_0}{cz_0}$.

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

​So, equation of normal at any point $P=(x_0,y_0,z_0)$ to the given ellipsoid is

$\frac{9(x-x_0)}{x_0}=\frac{4(y-y_0)}{y_0}=\frac{z-z_0}{z_0}$

When $x=0;$ Equation of normal becomes $-9=\frac{4(y-y_0)}{y_0}=\frac{z-z_0}{z_0}$

$\implies y=\frac{-5y_o}{4};z=-8z_o$

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

When $y=0; x=\frac{5x_o}{9};z=-3z_o$ $\implies Q_2=(\frac{5x_o}{9},0,-3z_o)$.

When $z=0;x=8x_o/9;y=3y_o/4$ $\implies Q_3=(\frac{8x_o}{9},\frac{3y_o}{4},0)$.

Hence $PQ_1 ​=\sqrt{(x_o)^2+(y_0-\frac{-5y_o}{4})^2+(z_0 + 8z_o)^2} = (\sqrt{16x_o^2+81y_o^2+36^2z_o^2})/4$

$PQ_2=\sqrt{(\frac{4x_o}{9})^2+(y_o)^2+(4z_o)^2} ​ = (\sqrt{16x_o^2​ +81y_0^2+36^2 z_0^2) }/9$

and $PQ_3 ​=\sqrt{(x_o - \frac{8x_0}{9})^2+(y_0-\frac{3y_o}{4})^2+z_0 ^2} = (\sqrt{16x_o^2+81y_o^2+36^2z_o^2})/36$ .

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

$\implies PQ_1 ​:PQ_2 ​:PQ_3 ​=9:4:1$