We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
193
1
avatar+846 

sorry if this is a long question.

 Oct 15, 2018
 #1
avatar+102422 
+3

We'll do (a)  last

 

(b) 

y^2  = x + 1    (1)

and

y  = m (x + 3)   square both sides

y^2  = m^2 (x + 3)^2  =  m^2 (x^2 + 6x + 9)  =  m^2x^2 + 6m^2x + 9m^2     (2)

 

Set the y^2 ' s   equal so that  (1)  = (2)

 

x + 1  = m^2x^2  + 6m^2x  + 9m^2

 

0 =  m^2x^2 + 6m^2x + 9m^2  - x  - 1

 

0 =  m^2x^2 + (6m^2 - 1)x  + (9m^2 - 1)

 

(c)

It can be shown [ using implicit differentiation ] that the slope of any tangent line to the  parabola  is =  1 / [2y]   = m

 

So

 

y = (m) (x + 3)

y =(1/ [2y]) ( x + 3)

2y^2 = x + 3         (3)

 

But

y^2  = x + 1   so

2y^2 = 2(x + 1)  (4)

 

Set  (3)  equal to (4)  and solve for x to find the x coordinate of the intersections

 

x + 3 = 2(x + 1)

x + 3 = 2x + 2

1  = x     [this is the x coordinate for the intersection of the line and the  parabola]

 

So....using this....we can find the y coordinates for the intersections

y^2 = x + 1

y^2  = 1 + 1

y^2  = 2

y = sqrt(2)  or  -sqrt(2) 

[ these are the y coordinates for the intersection of the line and the parabola ]

 

So....m = 1/ [2y] =  1/ [2 *±sqrt (2) ]  = 1 / [±sqrt(8)]

So...m^2  =  [ 1 / ±sqrt (8) ] ^2   =   1/8

 

 

(d)  The equations for the tangent lines  are

 

y = [1 / sqrt(8) ] (x + 3)    and

 

y = -(1/sqrt(8)] (x + 3)

 

And the intersection points are    [ 1, sqrt(2) ]  and  [ 1, - sqrt(2)  ]

 

(a)  Here's the graph :

 

https://www.desmos.com/calculator/jaa8mpc6d7

 

 

cool cool cool

 Oct 15, 2018
edited by CPhill  Oct 15, 2018

8 Online Users

avatar