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?
+506 Substitute \(y=3x+b\) into the first equation:
\(x^2+\frac{(3x+b)^2}{4}=1\\4x^2+(3x+b)^2=4\\4x^2+9x^2+6xb+b^2=4\\13x^2+6xb+b^2-4=0\)
In order for the line to be tangent to the ellipse, the above quadratic in terms of x must have exactly one solution, meaning that the discriminant has to be zero. Therefore,
\((6b)^2-4\cdot13\cdot(b^2-4)=0\\36b^2-52b^2+208=0\\b^2=13\\b=\boxed{\pm\sqrt{13}}\)
