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# Conic Question

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Circle gamma intersects the hyperbola y=1/x  at (1,1), (3, 1/3) and two other points. What is the product of the y coordinates of the other two points?

Guest Jul 17, 2018
#1
+19835
+2

Circle gamma intersects the hyperbola y=1/x  at (1,1), (3, 1/3) and two other points.

What is the product of the y coordinates of the other two points?

$$\text{Let P_1 = (x_1 = 1,\ y_1 = 1)} \\ \text{Let P_2 = (x_2 = 3,\ y_2 = \frac13 )} \\ \text{Center of the circle = (x_0,\ y_0)} \\ \text{Radius of the circle = r }$$

Formula of the circle:

$$\begin{array}{|lrcll|} \hline & (x-x_0)^2+(y-y_0)^2 &=& r^2 \\ \hline P_1(x_1,y_1): & (x_1 - x_0)^2 + (y_1 - y_0)^2 &=& r^2 \\ P_2(x_2,y_2): & (x_2 - x_0)^2 + (y_2 - y_0)^2 &=& r^2 \\ \hline (1)=(2): & (x_1 - x_0)^2 + (y_1 - y_0)^2 &=& (x_2 - x_0)^2 + (y_2 - y_0)^2 \\ & \Rightarrow \\ & ax_0+by_0 &=& c \\ & y_0 &=&\dfrac{c-ax_0}{b} \\ & &=& 3x_0-\dfrac{16}{3} \\ & \boxed{a=2(x_2-x_1) = 4 } \\ & \boxed{b=2(y_2-y_1) = -\dfrac{4}{3} } \\ & \boxed{c=x_2^2 +y_2^2-x_1^2-y_1^2=\dfrac{64}{9} } \\ \hline \end{array}$$

Circle intersects the hyperbola $$y=\dfrac{1}{x} \text{ or } x=\dfrac{1}{y}$$

$$\begin{array}{rcll} (x-x_0)^2+(y-y_0)^2 &=& r^2 \quad & | \quad r^2 = (x_1-x_0)^2+(y_1-y_0)^2 \\ (x-x_0)^2+(y-y_0)^2 &=& (x_1-x_0)^2+(y_1-y_0)^2 \quad & | \quad x=\dfrac{1}{y} \\ \left(\dfrac{1}{y}-x_0 \right)^2+(y-y_0)^2 &=& (x_1-x_0)^2+(y_1-y_0)^2 \\ \Rightarrow \\ \end{array} \\ \begin{array}{rcll} \boxed{ y^4-y^3\left(6x_0-\frac{32}{3}\right)+y^2\left(8x_0-\frac{38}{3}\right)-y\cdot2x_0+1= 0} \\ \end{array}$$

Examples calculated by WolframAlpha:

$$\begin{array}{|l|l|lr|l|} \hline x_0 & y_0 & y_3 & y_4 & y_3\cdot y_4 \\ \hline 1.4226 & -\dfrac{3.1966}{3} & -1.75493 & -1.70947 & 3.00000\ldots \\ 1.2 & -\dfrac{5.2}{3} & -4.06132 & -0.738675 & 2.99999\ldots \\ 1 & -\dfrac{7}{3} & -3-\sqrt{6} & \sqrt{6}-3 & 3 \\ 0.5 & -\dfrac{11.5}{3} & -8.65331 & -0.346688 & 2.99999\ldots \\ \hline \end{array}$$

I assume theproduct of the y coordinates of the other two points is 3

heureka  Jul 18, 2018
edited by heureka  Jul 18, 2018
#2
+92917
+1

Thanks Heureka, this one looks interesting.

Melody  Jul 18, 2018
#3
+19835
+1

Circle gamma intersects the hyperbola y=1/x  at (1,1), (3, 1/3) and two other points. What is the product of the y coordinates of the other two points?

$$\ldots$$continue

$$\begin{array}{rcll} \boxed{ y^4-y^3\left(6x_0-\frac{32}{3}\right)+y^2\left(8x_0-\frac{38}{3}\right)-y\cdot2x_0+1= 0} & (1)\\ \end{array}$$

$$\text{We have two roots y_1 = 1 and y_2 = \dfrac{1}{3}, } \\ \text{So formula (1) can be written as}:$$

$$\begin{array}{|rcll|} \hline && (y-y_1)(y-y_2)(y-y_3)(y-y_4) \quad | \quad y_1 = 1 \qquad y_2 = \dfrac{1}{3} \\ &=& (y-1)\left(y-\dfrac{1}{3}\right)(y-y_3)(y-y_4) \\ &=& (y^2-\dfrac13y - y+\dfrac13 )(y^2-yy_4-yy_3+y_3y_4 ) \\ &=& \left(y^2-\dfrac43y +\dfrac13 \right) \left(y^2-y(y_3+y_4)+y_3y_4 \right) \\ && \ldots \\ &=& y^4-y^3\left(\dfrac43 +y_3+y_4 \right) + y^2 \left( y_3y_4-\dfrac43(y_3+y_4 + \dfrac13 \right) - y \left(\dfrac43 y_3y_4-\dfrac13(y_3+y_4) \right) + \dfrac13y_3y_4 \\ \hline \end{array}$$

Compare:

$$\small{ \begin{array}{|rclclclclcl|} \hline && y^4 &-&y^3\left(6x_0-\frac{32}{3}\right) &+&y^2\left(8x_0-\frac{38}{3}\right) &-&y\cdot2x_0 &+& \color{red}1 \\ &=& y^4 &-&y^3\left(\dfrac43 +y_3+y_4 \right) &+&y^2 \left( y_3y_4-\dfrac43(y_3+y_4 + \dfrac13 \right) &-& y \left(\dfrac43 y_3y_4-\dfrac13(y_3+y_4) \right) &+& \color{red}\dfrac13y_3y_4 \\ \hline \end{array} }$$

$$\text{So  1 = \dfrac13 y_3y_4 \Rightarrow y_3y_4 = 3}$$

The product of the y coordinates of the other two points is 3

heureka  Jul 18, 2018
#4
+1

The equation of the circle can be written as

$$\displaystyle x^{2}+y^{2}+ax+by+c=0\;.$$

At intersections with the given hyperbola

$$\displaystyle \frac{1}{y^{2}}+y^{2}+\frac{a}{y}+by+c=0\;,$$

so,

$$\displaystyle y^{4}+by^{3}+cy^{2}+ay+1=0\;.$$

The product of the equations four roots will be the constant 1, two of the roots have a product of 1/3, so the product of the other two must be 3.

Guest Jul 18, 2018
#5
+1

How did you get it too the 4th degree?

Guest Jul 19, 2018
#6
0

Multiply the second equation by y^2, to clear the fractions.

Guest Jul 19, 2018