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In the coordinate plane, let F = (5,0). Let P be a point, and let Q be the projection of the point P onto the line x=16/5. The point traces a curve in the plane, so that \(\frac{PF}{PQ} = \frac{5}{4}\) for all points on the curve. Find the equation of this curve.(Enter it in standard form.) 

 Jan 28, 2019
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In the coordinate plane, let F = (5,0).

Let P be a point, and let Q be the projection of the point P onto the line \(x=\dfrac{16}{5}\).

The point traces a curve in the plane, so that  \(\dfrac{PF}{PQ} = \dfrac{5}{4}\) for all points on the curve.

Find the equation of this curve.

(Enter it in standard form.) 

 

The equation of this curve is a hyperbola : \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\)

The focus is the Point \(F = (5,0) =(c,0)\)  so \(c = 5 \).

 

\(\mathbf{a=\ ?}\)

\(\begin{array}{|rcll|} \hline \dfrac{PF}{PQ} = \dfrac{c}{a} &=& \dfrac{5}{4} \\\\ \dfrac{c}{a} &=& \dfrac{5}{4} \quad | \quad c = 5 \\\\ \dfrac{5}{a} &=& \dfrac{5}{4} \\\\ \dfrac{1}{a} &=& \dfrac{1}{4} \\\\ \mathbf{a} &\mathbf{=}& \mathbf{4} \\ \hline \end{array} \)

 

\(\mathbf{b=\ ?}\)

\(\begin{array}{|rcll|} \hline b^2 &=& c^2-a^2 \quad | \quad c = 5,~ a= 4 \\ b^2 &=& 5^2-4^2 \\ b^2 &=&25-16 \\ b^2 &=& 9 \\ \mathbf{b} &\mathbf{=}& \mathbf{3} \\ \hline \end{array}\)


check:
\(\begin{array}{rcl} \dfrac{16}{5} &=& \dfrac{a^2}{c} \\\\ &=& \dfrac{4^2}{5} \\\\ &=& \dfrac{16}{5}\ \checkmark \\ \end{array}\)

 

\(\text{Let $ P=(x_p,y_p) $ }\)

 

The equation of this curve is : \(\mathbf{\dfrac{x_p^2}{4^2} - \dfrac{y_p^2}{3^2} = 1} \)

 

 

laugh

 Jan 28, 2019
edited by heureka  Jan 28, 2019

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