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The equation of an ellipse is x^2+y^2/4=1. The equation of a line is y=2x+b. For what values of b is the line tangent to the ellipse?

 
 Nov 22, 2020
 #1
avatar+112375 
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x^2 + y^2/4  = 1      using implicit differentiation  we have that

 

2x + (1/2)yy'  = 0

 

y'  =  -2x / (1/2 y)

 

y'  =  -4x/ y

 

Yhe slope of the line  = 2

 

So....we want to find where

 

-4x / y =  2

 

-4x  = 2y

 

-2x  =  y

 

Sub this back into the equation of the ellipse for y

 

x^2   +  (-2x)^2/4  = 1

 

x^2  + 4x^2/4  = 1

 

2x^2  =  1

 

x^2  =   ±√(1/2)

 

When  x = √(1/2),  y =  -2√(1/2)

When x = -√(1/2) ,  y = 2√(1/2)

 

Subbing into the equation of the line

 

 -2√(1/2)  = 2√(1/2)  + b   →  b =  - 4√(1/2)

And

2√(1/2)  = -2√(1/2)  + b →  b = 4√(1/2)

 

See the graph here  :  https://www.desmos.com/calculator/7ifp0ajbri

 

 

cool cool cool

 
 Nov 22, 2020
 #2
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0

Thanks CPHill! I already solved this problem lol but anyone is welcome to try it!

 
 Nov 22, 2020
 #3
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+1

Hold on... I got 2sqrt(2). I did a different method. 

 
 Nov 22, 2020
 #4
avatar+112375 
+1

Notice  that   4sqrt (1/2)  =    ±  4 (1) / sqrt (2)  =  4/sqrt (2)

 

Multiply  top/bottom  by  sqrt (2)   and we get

 

 ±  4sqrt (2) / 2 =   ±  2 sqrt (2)  =  b

 

Sorry that I didn't simplify it  more  !!!

 

cool cool cool

 
 Nov 22, 2020
edited by CPhill  Nov 22, 2020

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