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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
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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   Nov 22, 2020
#2
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Thanks CPHill! I already solved this problem lol but anyone is welcome to try it!

Nov 22, 2020
#3
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Hold on... I got 2sqrt(2). I did a different method.

Nov 22, 2020
#4
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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  !!!   Nov 22, 2020
edited by CPhill  Nov 22, 2020