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# Conic section

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1 sorry if this is a long question.

Oct 15, 2018

### 1+0 Answers

#1
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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   Oct 15, 2018
edited by CPhill  Oct 15, 2018