Answer to Question #91591 in Analytic Geometry for Dinah

Question #91591
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)
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Expert's answer
2019-07-11T08:26:47-0400


2.1 Let P(xP,yP),Q(xQ,yQ).P(x_P, y_P), Q(x_Q, y_Q). Then


xP=yP,yP=49xP,xP>0,yP>0x_P=y_P, y_P={4 \over 9x_P}, x_P>0, y_P>0xQ=yQ,yQ=49xQ,xQ<0,yQ<0x_Q=y_Q, y_Q={4 \over 9x_Q}, x_Q<0, y_Q<0

yP=49=23=xPy_P=\sqrt{4 \over 9}={2 \over 3}=x_PyQ=49=23=xQy_Q=-\sqrt{4 \over 9}=-{2 \over 3}=x_Q

P(23,23),Q(23,23)P({2 \over 3}, {2 \over 3}), Q(-{2 \over 3}, -{2 \over 3})

2.2 Let S(xS,yS),R(xR,yR).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


xS=0,yS=yP=23x_S=0, y_S=y_P={2 \over 3}xR=xP=23,yR=0x_R=x_P={2 \over 3}, y_R=0

S(0,23),R(23,0)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(23,23),radius=23P({2 \over 3}, {2 \over 3}), radius={2 \over 3}(xxP)2+(yyP)2=(radius)2(x-x_P)^2+(y-y_P)^2=(radius)^2


Write down the equation of the circle 


(x23)2+(y23)2=49(x-{2 \over 3})^2+(y-{2 \over 3})^2={4 \over 9}

2.4 The equation of the line through S and R



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




x0230=y23023{x-0 \over {2 \over 3}-0}={y-{2 \over 3} \over 0-{2 \over 3}}

x=y+23x=-y+{2 \over 3}

The equation of the line through T, S and R


y=x+23y=-x+{2 \over 3}

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


xT=xQ=23x_T=x_Q=-{2 \over 3}

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


yT=xT+23y_T=-x_T+{2 \over 3}yT=(23)+23=43y_T=-(-{2 \over 3})+{2 \over 3}={4 \over 3}

T(23,43)T(-{2 \over 3}, {4 \over 3})Q(23,23)Q(-{2 \over 3}, -{2 \over 3})

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


TQ=yTyQTQ=|y_T-y_Q|

TQ=43(23)=2TQ=|{4 \over 3}-(-{2 \over 3})|=2TQ=2unitsTQ=2 units


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Comments

Assignment Expert
12.07.19, 16:08

Dear Dinah, You are welcome. We are glad to be helpful. If you liked our service, please press a like-button beside the answer field. Thank you! We try our best to provide detailed solutions of questions.

Dinah
11.07.19, 23:11

This site is amazing please keep the good work up,but please give more details of the solutions for better understanding. thank you

Assignment Expert
11.07.19, 15:33

Dear Siphiwe. The solution of the question has already been published.

Siphiwe
11.07.19, 13:23

Please help with 2,2...i got the coordinate as (3/2:3/2) for 2,1

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