Question

the ratio of distance of P from (8,0) to distance from x=25/8 is 8:5 then the locus of point is​

Answer 1

Locus of a point

Given: the ratio of distance of $P$ from $(8,0)$ to distance from $x=\frac{25}{8}$ is $8:5$

To find: the locus of the point $P$

Solution:

Let the coordinates of $P$ be $(a,b)$

• The distance of $P$ from $(8,0)$ is
• $=\sqrt{(a-8)^{2}+b^{2}}$

• and the distance of $P$ from $x=\frac{25}{8}$ is
• $=|a-\frac{25}{8}|$

By the given condition,

$\quad \sqrt{(a-8)^{2}+b^{2}}:|a-\frac{25}{8}|=8:5$

$\Rightarrow \frac{\sqrt{(a-8)^{2}+b^{2}}}{|a-\frac{25}{8}|}=\frac{8}{5}$

$\Rightarrow \frac{64\{(a-8)^{2}+b^{2}\}}{(8a-25)^{2}}=\frac{64}{25}$

$\Rightarrow 1600(a^{2}-16a+64+b^{2})=64(64a^{2}-400a+625)$

$\Rightarrow 25(a^{2}-16a+64+b^{2})=64a^{2}-400a+625$

$\Rightarrow 25a^{2}-400a+1600+25b^{2}=64a^{2}-400a+625$

$\Rightarrow 39a^{2}-25b^{2}=975$

$\Rightarrow \frac{a^{2}}{25}-\frac{b^{2}}{39}=1$

Hence the required locus of $P$ is

• $\frac{x^{2}}{25}-\frac{y^{2}}{39}=1$, which is a hyperbola.