Find the locus of a point which moves such that the ratio of its distance from the point $(-5,0)$ to its distance from the line $5x+ 16 = 0$ is $5/4$.
I was trying this question, and I got my answer $x= -16/5$. I don't know how to combine the equation, but I know that its eccentricity is $5/4$. I was not able to find the locus of moving point. I don't know from where I have to start.
Thanks to anybody who can help me.
On
Let $P(x,y)$ be the point Its distance from $A(-5,0)$ is
$PA=\sqrt{(x+5)^2+y^2}$
its distance from the line $x=-\dfrac{16}{5}$ is $PH=\left|x+\dfrac{16}{5}\right|$
$PA=\dfrac{5}{4}\,PH \to PA^2=\dfrac{25}{16}\,PH^2$
so we have
$(x+5)^2+y^2=\dfrac{25}{16}\left(x+\dfrac{16}{5}\right)^2$
that is the hyperbola
$\dfrac{x^2}{16}-\dfrac{y^2}{9}=-1$
Take a generic point $P=(x,y)$. Write down an expression $f(x,y)$ for the distance of $P$ from line $x=-16/5$. Write down a second expression $g(x,y)$ for the distance of $P$ from point $(-5,0)$. The equation $$ g(x,y)={5\over4}f(x,y) $$ represents the locus. You only need to simplify it a bit.