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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?

Feng
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funfun
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    I think the more important question is: How would you arrive at the result by calculations? I, for one, don’t understand what you mean by “I answered through the Cartesian plane.” If you mean that you solved this graphically, how do you know that the values are exactly $\pm\sqrt5?$ instead of, say, $\pm9/4$? Those only differ by a couple of hundreths, which is likely well within the margin of error of your drawing or screen resolution. – amd Aug 04 '19 at 00:00

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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 $5-n^2=0$

Dr. Sonnhard Graubner
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