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The equation of an ellipse is x^2+y^2/4=1. The equation of a line is y=3x+b. For what values of b is the line tangent to the ellipse?

 Feb 18, 2022
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Using implicit differentiation  we can find the slope at any point on the  ellipse

 

2x  + (1/4)(2))yy'  =  0

y' =     -2x / [ (1/2) y ]    =   -4x / y

 

We know the slope of the line   =3

So

 

-4x / y  =  3

-4x = 3y

(-4/3)x = y           

 (16/9)x^2 = y^2       sub this  back into the  equation of the ellipse

 

x^2  + (1/4)(16/9)x^2  =  1

x^2 + (4/9)x^2  = 1

(13/9)x^2  = 1

x^2  =  (9/13)              take both roots

x=  +3/sqrt 13     and x  = -3/sqrt 13

 

When x = 3/sqrt 13 , y=  (-4/3)x  = (-4/3) (3/sqrt 13) =  -4/sqrt 13

When x = -3/5 , y =  (-4/3)x = -(4/3)(-3/sqrt 13)  = 4/sqrt 13

 

So  the line is tangent to the ellipse at   ( 3/sqrt 13, -4/sqrt 13)   and  (-3/sqrt 13, 4/sqrt 13)

 

Since b is the y interecept in both cases let  (0, b)  be the b we are looking for

 

And we can use the slope formula to find b  in  each case

 

( b - -4/sqrt 13) / ( 0 - 3/sqrt 13)  = 3                            

(b + 4/sqrt 13) / ( -3/sqrt 13) = 3

b + 4/sqrt 13  = -9sqrt 13

b  = -13/sqrt 13 =  -sqrt 13

 

And by symmetry,  b = sqrt 13  is the other value for b

 

Here's a graph : https://www.desmos.com/calculator/kaztwjaevr

 

 

cool cool cool

 Feb 18, 2022

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