Question

A point $P$ moves on a plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$. A plane through $P$ and perpendicular to $OP$ meets the coordinate axes in $A,B$ and $C$. If the planes through $A,B$ and $C$ parallel to the planes $x=0,y=0$ and $z=0$ intersect in $Q$, find the locus of $Q$.

• A. $\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$
• B. $\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$
• C. $\frac{1}{x^2}-\frac{1}{y^2}-\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$
• D. $x^2+y^2+z^2=ax+by+cz$

Answer 1

Answer: (A)

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$

Let $(\alpha ,\beta ,\gamma )$ be the coordinate of $Q$

As the planes through the point $A,B,C$ parallel to the coordinate planes, meet at the point $Q$, the co-ordinate of $A,B,C$ are respectively $(\alpha ,0,0),(0,\beta ,0)$ and $(0,0,\gamma )$.

Then the equation of the plane $ABC$ is $\frac{x}{\alpha }+\frac{y}{\beta }+\frac{z}{\gamma }=1$ ...$\left(1\right)$

Let $P(a^{\prime },b^{\prime },c^{\prime })$ be any point on the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$ ...$\left(2\right)$

Let $O$ denote the origin. The plane through $P$, perpendicular to $OP$, meets the axes in $A,B,C$ is $Op$ lies on the plane $\left(1\right)$ and $OP$ is perpendicular to the plane $\left(1\right)$

So, $\frac{a^{\prime }}{\frac{1}{\alpha }}+\frac{b^{\prime }}{\frac{1}{\beta }}+\frac{c^{\prime }}{\frac{1}{\gamma }}=\lambda$(say) ...$\left(3\right)$

As $P$ lies on the both planes $\left(1\right)and\left(2\right)$

$\frac{a^{\prime }}{a}+\frac{b^{\prime }}{b}+\frac{c^{\prime }}{c}=1$ ...$\left(4\right)$

and $\frac{a^{\prime }}{\alpha }+\frac{b^{\prime }}{\beta }+\frac{c^{\prime }}{\gamma }=1$ ...$\left(5\right)$

Using $\left(3\right)in\left(4\right)and\left(5\right)$, we get

$\frac{\lambda }{a\alpha }+\frac{\lambda }{b\beta }+\frac{\lambda }{c\gamma }=1$ ...$\left(6\right)$

and $\frac{\lambda }{{\alpha }^2}+\frac{\lambda }{{\beta }^2}+\frac{\lambda }{{\gamma }^2}=1$ ...$\left(7\right)$

Eliminating $\lambda$ between $\left(6\right)$ and $\left(7\right)$, we obtain

$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{1}{{\gamma }^2}=\frac{1}{a\alpha }+\frac{1}{b\beta }+\frac{1}{c\gamma }$

Hence the required locus of $(\alpha ,\beta ,\gamma )$

$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{ax}+\frac{1}{by}+\frac{1}{cz}$