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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.

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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

$\dfrac{x^2}{16}-\dfrac{y^2}{9}=-1$

enter image description here

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  • $\begingroup$ thanks a lot @ raffaele $\endgroup$ – Lily Aug 7 '17 at 12:41
  • $\begingroup$ Please, vote me as best answer :) $\endgroup$ – Raffaele Aug 7 '17 at 12:48
  • $\begingroup$ i dont know how to vote where is the vote option $\endgroup$ – Lily Aug 7 '17 at 13:10
  • $\begingroup$ Near my answer, top left there is a double arrow up to vote up and down to vote down. The original poster, you in this case, can choose the best answer too $\endgroup$ – Raffaele Aug 7 '17 at 13:22
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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.

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