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?
+128408 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
