The actual values $n$ for which the line (t) $y = x + n$ is tangent to the ellipse of equation $2x^2 + 3y^2 = 6$ are equal to:
note: I answered through the Cartesian plane, the answers are $-\sqrt5$ and $\sqrt5$. How would you arrive at the result by calculations?
Hint: Plug the equation of the line into the equation of the ellipsis $$2x^2+3(x+n)^2=6$$ solve this equation for $x$ and set the discriminant equal to Zero to determe $n$ For your work: $$x_1=\frac{1}{5} \left(-\sqrt{6} \sqrt{5-n^2}-3 n\right)$$ $$x_2=\frac{1}{5} \left(\sqrt{6} \sqrt{5-n^2}-3 n\right)$$ so we get $5-n^2=0$