+0

# help

0
240
1

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

#1
+8

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 : $$\mathbf{\dfrac{x_p^2}{4^2} - \dfrac{y_p^2}{3^2} = 1}$$  Jan 28, 2019
edited by heureka  Jan 28, 2019