Question

A variable line through the pt. $(\frac{6}{5},\frac{6}{5})$ cuts the Coordinate axis at the pt. A & B. If the P divides AB internally in the ratio $2:1$. Find locus of the point.

• A. 2xy= 2(5x+y)
• B. 5xy= 2(2x+y)
• C. 3xy= 12x
• D. none

Answer 1

Answer: (B)

5xy= 2(2x+y)

Let equation of line by $\to \frac{x}{a}+\frac{y}{b}=1$ [When line cut x axis a and y axis at b]

$A(a,0)$

$B(0,b)$ line passes through $\frac{6}{5}+\frac{6}{5}$

$\frac{6}{5a}+\frac{6}{5b}=1$

$\frac{1}{a}+\frac{1}{b}=\frac{5}{6}\to \left(1\right)$

Using section formula

$x=\frac{m{x}_2+n{x}_1}{m+n}$ $y=\frac{m{y}_2+n{y}_1}{m+n}$

If point $(x,y)$ divides line joining the two points $(x_1,y_1)$ and $(x_2,y_2)$

then x and y are given by above formula

$x=\frac{2\ast 0+a}{3}$ $y=\frac{2\ast b+0}{3}$

$x=\frac{a}{3}$ $y=\frac{2b}{3}$

$\frac{1}{a}=\frac{1}{3x}\to \left(2\right)$ $\frac{1}{b}=\frac{2}{3y}\to \left(3\right)$

Add 2 and 3

$\frac{1}{a}+\frac{1}{b}=\frac{1}{3x}+\frac{2}{3y}$

$\frac{5}{6}=\frac{1}{3}[\frac{1}{x}+\frac{2}{y}]$

$5xy=2y+4x$

$5xy=2(2x+y)$ Ans