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

As per the given condition in the question,

The equation of the ellipse,

$\dfrac{x^2}{9}+\dfrac{y^2}{4}+z^2=1$

now compare the above the general equation of the ellipse,

$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$

Let the parametric point $P(a\sin\theta\cos\phi, b\cos\theta\sin\phi, c\cos\theta)$

a=3, b=2, c=1

now putting $(a\sin\theta\cos\phi, 0, 0)$,

We know from the general equation,

we know that from the general equation,

$x=-\dfrac{a}{a^2-b^2-c^2}=\dfrac{9}{-4}$

similarly $y=-\dfrac{b}{a^2-b^2-c^2}=\dfrac{4}{-4}$

similarly $z=\dfrac{c}{a^2-b^2-c^2}=\dfrac{-1}{4}$

So, distance $PQ_1=\sqrt{(x-0)^2+(0-0)^2+(0-0)^2}=\dfrac{9}{4}$

Similarly $PQ_2=\dfrac{4}{4}$

$PQ_3=\dfrac{1}{4}$

Hence the required ratio is 9:4:1