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 #1
avatar+124676 
+3

There is only one value of k for which the line x = k intersects the graphs of  y = x^2 + 6x + 5 and y = mx + b at two points which are exactly 5 units apart. If the line y = mx + b passes through the point (1, 6), and b ≠ 0, find the equation of the line. Enter your answer in the form "y = mx + b".

 

For convenience  let    the  vertical  line be  x  = r

Then the point on the parabola  that this line intersects =  (r , r^2 + 6r + 5)

 

Call the point on the parabola   that the line of interest intersects  =  (q , q^2 + 6q + 5)

 

The slope of this line =   (6 - [ q^2 + 6q + 5 ])  / ( 1 - q)  =    ( 1 - q^2 - 6q) / (1 -q)  =  (q^2 + 6q - 1) / ( q - 1)  =  m

 

You  have correctly figured out the  6 = m + b

So 6 - m = b

So  6 -  (q^2 + 6q - 1) / (q - 1)   = b

So  [ 6 (q -1) -  (q^2 + 6q - 1) ]  / (q -1)  = b

So  [ q^2 + 5  ] / (1 - q)  =  b 

 

Then   the point   (0, b)  =    [ 0 , q^2 + 5 ] / [1 -q)]   is on  the line

The slope between   this  point and (1,6)  =      [  6 - q^2 - 5 ] / [  (1-q)  ]  =  (1 - q^2] / (1 - q)  =  1 + q

Equating slopes

 

( 1 + q)  = ( q^2 + 6q -1) / (q -1)   

q^2 - 1  = q^2 + 6q - 1

0 = 6q        →   q =  0

 

So   the line of interest intersects the parabola at    [ 1-q , q^2 + 5  ]  =  ( 0,5)

So  m  = slope between (0, 5) and (6, 1)  =    (6 -5) / (1 -0) =  1

And 

b = 5

Our line  is   y = x + 5    { verify that   (1,6)   is on this  line }

 

When  x = r

Then y =  r + 5

 

And we need the distance  between   (r , r+5  )    and  ( r,  r^2 + 6r + 5)   to equal   5

 

So

 

sqrt  [   (r - r)^2 + [ (r +5) - (r^2 + 6r + 5) ]^2  ] =  5

 

A little tricky to solve....but..solving this  for the  r  we need produces  

r ≈ .8541

So the vertical line is  x =  .8541

And y  ≈

(.8541)^2  + 6(.8541) + 5  ≈ 10.8541  

 

So   ( .8541, 10.8541)  =  "B"  on the  graph  = the place where the vertical line crosses the parabola

And

(r , r + 5)  =   ( .85410 , 5.8541) = "B"  on the  graph  where the vertical line intersects y = x + 5

 

Here's a picture :

 

 

 

cool cool cool

21 ago. 2022
 #1
avatar+124676 
+2

See below

 

 

We can  construct two circles with equations

x^2 + y^2   =4

(x -3)^2 + y^2  = 1

These circles are tangent at P = (2,0)

 

We can also construct similar triangles   ABD  and ACE   where  points D and E are  the points of tangency of our line of interest and  the two circles

 

We can find point  "A"  as follows

Let   a =  the distance from  A  to the smaller circle

AB = (radius of smaller circle + a) = 1 + a

AC = ( radius of larger circle + a)  =   4 + a

BD = 1

CE = 2

 

 

So  we  have this relationship  in these similar triangles

AB / AC  =  BD / CE

(1 + a) / ( 4 + a)  =   1/2     cross-multiply

2 (1 + a)  =  1 (4 + a)

2 + 2a  = 4 + a

2a  - a =  4  - 2

a =  2

 

So  AB  =   1 + 2   = 3

So CA =  (2 + 1 + 3)  = 6

So point A has the coordinates (6,0)

 

Now since  triangle ABD is right with hypotenuse AB,  then  AD  = sqrt ( AB^2 - BD^2)  = sqrt (3^2 -1^2) =sqrt (8)

The  tangent  of  angle  BAD  = BD / AD  =   1 /sqrt (8)   = the slope of our line

So the equation of our line  going through (6,0) is 

y  =   (1/sqrt (8)) ( x - 6)

Multiply through by  sqrt 8

sqrt(8)y  = x -6

 

Putting  in standard form

x - sqrt (8) y - 6  = 0

 

The distance  from point  P  = (2,0)   to this line  is given  by

 

 

          abs  [  2 - sqrt (8)(0) - 6 ]               abs (-4)               4

      _______________________  =      _________  =     ___    =   PF

          sqrt  [ 1^2  + [ sqrt (8)]^2                sqrt (9)               3

 

So   m + n  = 7

 

cool cool cool

19 ago. 2022