+0

# The hyperbola (y-3)^2/5 -(x-2)^2/16=1 (the equation is in x^2/a^2-y^2/b^2=1 Intersects the line at x=8 at (8,y1) and (8,y2) find y1+y2

0
162
1

The hyperbola (y-3)^2/5 -(x-2)^2/16=1 (the equation is in x^2/a^2-y^2/b^2=1
Intersects the line at x=8 at (8,y1) and (8,y2) find y1+y2

Feb 17, 2021

### 1+0 Answers

#1
+1

(y - 3)^2  /  5  -  (x - 2)^2  / 16  = 1

Since it  intersects the  line  x  = 8.....just  sub this value of  x  into  the  equation

(y - 3)^2  / 5    - ( 8  -2)^2 / 16   = 1     simplify

Multiply  through  by  the LCM  of  5, 16   =  80

16 ( y - 3)^2    -  5 ( 8 - 2)^2  =  80       simplify

16 ( y -3)^2  - 5*6^2    = 80

16 ( y - 3) ^2  =  180  +  80

16 ( y - 3)^2   =  260    divide  both sides  by  16

(y - 3)^2   = 260/16      take  both roots

y - 3   =  sqrt ( 260/16)       y  =   3  + 2sqrt (65)  /4  =    3 + sqrt (65) / 2  =  y1

and

y  -  3  =  - sqrt (260/16)     y =  3   - 2 sqrt (65) /4  =   3  - sqrt (65)  /2  =  y2

So

y1  +  y2  =       6   Feb 17, 2021