Solve for the value where $k$ is a tangent to the ellipse.

Question

For the ellipse $4x^2 + 9y^2 = 36$ and the function $2x + 3y = k$ find the values of $k$ that are a tangent to the ellipse.

I squared the function and got $ 4x^2 + 9y^2 = k^2$

When I solve it I get $k = \pm\sqrt 6$ don't know where I went wrong, as the answer key says it's $\pm2\sqrt6$ .

Any help for why I am missing the "$2$" at the end of the equation would be appreciated. Thank you.

Answer 1

The given line is $y=-\frac 23 x+\frac k3$, substitute this in original and you will get the point of intersection of line and ellipse, but for any line to be tangent only one point of intersection is there thus the discriminant of quadratic you will get should be $0$.

Answer 2

$$(a+b)^2=a^2+b^2+2ab$$

As Jonathan Richard Lombardy already stated this is what you're missing.

Answer 3

$u = 2x, v = 3y\\ u^2 + v^2 = 36\\ u + v = k$

Now we have a circle and not an ellipse, and circles are easier to work with.

the tangent to the circle is perpendicular to the radius. the point of tangency is then on the line $u = v$

$u,v = \pm 6\frac {\sqrt{2}}{2}\\ k = \pm 6{\sqrt{2}}$

Now that doesn't fit your answer key, either.