Question

A plane passes through a fixed point $(a,b,c)$ and cuts the axes in $A,B,C$. The locus of a point common to the plane through, $A,B$ and $C$ and parallel to coordinate plane :

• A. $ayz+bzx+cxy=2xyz$
• B. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=1$
• C. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$
• D. $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=3$

Answer 1

The correct option is C $\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$

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Consider the equation of sphere $OABC$ through $\left(0,0,0\right)$ is

$x^2+y^2+z^2+2ux+2vy+2wz=0.............\left(1\right)$

now $\left(1\right)$ meets x-axis where $y=0$ and $z=0$

put, $y=2=0$ in $\left(1\right)$ then we find

$\begin{array}{c}{x}^2+2ux=0\\ \Rightarrow x=-2u\end{array}$

So, $\left(1\right)$ meets x-axis at $A\left(-2u,0,0\right)$

Similarly $\left(1\right)$ meet y-axis and z-axis at $B\left(0,-2,0\right)$ and $C\left(0,0,-2w\right)$

Now equation of plane$ABC$ is

$\begin{array}{c}\frac{x}{-2u}+\frac{y}{-2v}+\frac{z}{-2w}=1\\ \Rightarrow \frac{x}{-u}+\frac{y}{-v}+\frac{z}{-w}=2\end{array}$

since it passes through $(A,B,C)$ then

$\frac{a}{-u}+\frac{b}{-v}+\frac{c}{-w}=2.............\left(2\right)$

if $\left(x_1,y_1,z_1\right)$ is the center of the sphere $\left(1\right)$ then

$x_1=-u;y_1=-v;z_1=-w$

now from $\left(2\right)$ we get

$\frac{a}{x_1}+\frac{b}{y_1}+\frac{c}{z_1}=2$

So locus of the center $\left(x_1,y_1,z_1\right)$ of sphere is

$\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=2$