Question

find the locus of the point which moves show that ratio of its distance from the point (-5,0) and (5,0) is 2:3

Answer 1

Answer:

Step-by-step explanation:

Let the required point be (h, k)

According to the question,

$\tt{\dfrac{\sqrt{\left(k-0\right)^2+\left(h+5\right)^2}}{\sqrt{\left(k-0\right)^2+\left(h-5\right)^2}}=\dfrac{2}{3}}$

Squaring both sides,

$\tt{\left(\dfrac{\sqrt{\left(k-0\right)^2+\left(h+5\right)^2}}{\sqrt{\left(k-0\right)^2+\left(h-5\right)^2}}\right)^2=\left(\dfrac{2}{3}\right)^2}$

$\tt{\implies\dfrac{\left(k-0\right)^2+\left(h+5\right)^2}{\left(k-0\right)^2+\left(h-5\right)^2}=\dfrac{4}{9}}$

$\tt{\implies\dfrac{k^2+\left(h+5\right)^2}{k^2+\left(h-5\right)^2}=\dfrac{4}{9}}$

$\tt{\implies\dfrac{k^2+h^2+10\,h+25}{k^2+h^2-10\,h+25}=\dfrac{4}{9}}$

$\tt{\implies9\left(k^2+h^2+10\,h+25\right)=4\left(k^2+h^2-10\,h+25\right)}$

$\tt{\implies9\,k^2+9\,h^2+90\,h+225=4\,k^2+4\,h^2-40\,h+100}$

$\tt{\implies9\,k^2+9\,h^2+90\,h+225-4\,k^2-4\,h^2+40\,h-100=0}$

$\tt{\implies9\,h^2-4\,h^2+9\,k^2-4\,k^2+90\,h+40\,h+225-100=0}$

$\tt{\implies5\,h^2+5\,k^2+90\,h+40\,h+125=0}$

$\tt{\implies\,h^2+k^2+18\,h+8\,h+25=0}$

$\tt{\implies\,h^2+k^2+26\,h+25=0}$

$\boxed{\bf{Required\,\,\,Locus\,:\,\,x^2+y^2+26x+25=0}}$