Answer to Question #91591 in Analytic Geometry for Dinah

Question

The sketch shows the hyperbola defined by y the straight line defined by y ; a circle with centre at P, touching the r-axis and y-axis at R and S, respectively; and the straight line through the points T, S and R. The line joining T and Q is parallel to the y-axis. 2.1 Determine the coordinates of P and Q (5) 2.2 Determine the coordinates of S and R (2) 2.3 Find the radius of the circle, and write down the equation of the circle (3) 2.4 Determine the equation of the line through T, S and R. (5) 2.5 Calculate the length of the line TQ (5)

Answer 1

2.1 Let $P(x_P, y_P), Q(x_Q, y_Q).$ Then

$$x_P=y_P, y_P={4 \over 9x_P}, x_P>0, y_P>0$$ $$x_Q=y_Q, y_Q={4 \over 9x_Q}, x_Q<0, y_Q<0$$

$$y_P=\sqrt{4 \over 9}={2 \over 3}=x_P$$ $$y_Q=-\sqrt{4 \over 9}=-{2 \over 3}=x_Q$$

$$P({2 \over 3}, {2 \over 3}), Q(-{2 \over 3}, -{2 \over 3})$$

2.2 Let $S(x_S, y_S), R(x_R, y_R).$ Since a circle with centre at P touches the x-axis and y-axis at R and S, respectively, then

$$x_S=0, y_S=y_P={2 \over 3}$$ $$x_R=x_P={2 \over 3}, y_R=0$$

$$S(0, {2 \over 3}), R({2 \over 3}, 0)$$

2.3 A circle with centre at P touches the x-axis and y-axis at R and S, respectively.

$$P({2 \over 3}, {2 \over 3}), radius={2 \over 3}$$ $$(x-x_P)^2+(y-y_P)^2=(radius)^2$$

Write down the equation of the circle 

$$(x-{2 \over 3})^2+(y-{2 \over 3})^2={4 \over 9}$$

2.4 The equation of the line through S and R

$${x-x_S \over x_R-x_S}={y-y_S \over y_R-y_S}$$

$${x-0 \over {2 \over 3}-0}={y-{2 \over 3} \over 0-{2 \over 3}}$$

$$x=-y+{2 \over 3}$$

The equation of the line through T, S and R

$$y=-x+{2 \over 3}$$

2.5 Let $T(x_T, y_T)$. Since the line joining T and Q is parallel to the y-axis, then

$$x_T=x_Q=-{2 \over 3}$$

Since the point T lies on the line through T, S and R, then

$$y_T=-x_T+{2 \over 3}$$ $$y_T=-(-{2 \over 3})+{2 \over 3}={4 \over 3}$$

$$T(-{2 \over 3}, {4 \over 3})$$ $$Q(-{2 \over 3}, -{2 \over 3})$$

Since $x_T=x_Q,$ then the length of the line TQ is

$$TQ=|y_T-y_Q|$$

$$TQ=|{4 \over 3}-(-{2 \over 3})|=2$$ $$TQ=2 units$$