Question

A variable plane is at a constant distance $2p$ from the origin and meets the coordinate axes in point $A$, $B$ and $C$ respectively. Through these points, planes are drawn parallel to the coordinates plane. Find the locus of their point of intersection

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{4p^2}$
• B. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{p^2}$
• C. $\frac{16}{x^2}+\frac{16}{y^2}+\frac{16}{z^2}=\frac{1}{p^2}$
• D. $xyz=p^3$

Answer 1

The correct option is A $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{4p^2}$

Lets Consider normal to plane be $a\hat{i}+b\hat{j}+c\hat{k}$
Plane equation of unit normal vector $\hat{n}$ and perpendicular distance between origin and plane is $d$ then plane equation is $\hat{n}.(x\hat{i}+y\hat{j}+z\hat{k})=d$
Then plane equation would be $(a\hat{i}+b\hat{j}+c\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=2p\sqrt{a^2+b^2+c^2}$
it meet coordinate axis at $x=\frac{2p\sqrt{a^2+b^2+c^2}}{a},y=\frac{2p\sqrt{a^2+b^2+c^2}}{b}=z=\frac{2p\sqrt{a^2+b^2+c^2}}{c}$
Three plane parellel to coordinate plane intersect at point whose coordinate value will be equal to to its intercept
since $\frac{1}{x^2}=\frac{a^2}{4p^2(a^2+b^2+c^2)}$
Like this for $\frac{1}{y^2}=\frac{b^2}{4p^2(a^2+b^2+c^2)},\frac{1}{z^2}=\frac{b^2}{4p^2(a^2+b^2+c^2)}$
Hence
$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{(a^2+b^2+c^2)}{4p^2(a^2+b^2+c^2)}=\frac{1}{4p^2}$