Show that the line $x+y=q$ intersects the ellipse $x^2-2x+2y^2=3$ at two different points if $q^2<2q+5$

Question

What should be the proper way to answer this question?

My working is as follows, but is it the right way to answer the question, since it seems like I am only showing that $q^2<2q+5$?

$x+y=q \tag{1}$

$x^2-2x+2y^2=3 \tag{2}$

From $(1)$, $y=q-x\tag{3}$

Substitute $(3)$ into $(2)$,

$\begin{align}x^2-2x+2(q-x)^2&=3\\x^2-2x+2(q^2-2qx+x^2)&=3\\3x^2-(2+4q)x+2q^2-3&=0\end{align}$

Since the line intersects the ellipse at two different points, therefore $b^2-4ac>0$

$(-2-4q)^2-4(3)(2q^2-3)>0$

$4+16q+16q^2-24q^2+36>0$

$8q^2-16q-40<0$

$q^2-2q-5<0$

$q^2<2q+5$

Does my working answer the quesiton, or are there more accurate approaches to solve this question?

Answer 1

If $y=x-q$ intersects the ellipse $$x^2-2x-2y^2=3 \implies x^2-2x+2(x-q)^2=3$$ Then the roots of this quadratic have to be real. So we demand $B^2 >4AC$ in the simplified quadratic, $$3x^2-(4q+2)x+2q^2-3=9 \implies q^2 <2q+5$$

Answer 2

Your calculations are correct. Here you can find also for what values of $q$ the line $x+y=q$ and the ellpise $x^2-2x+2y^2=3$ have two points in common.

First, you can rewrite the equation of the ellipse in the form: $$(x-1)^2+2y^2=4$$ Now, substituing $y=q-x$, you have: $$(x-1)^2+2(q-x)^2=4\leftrightarrow 3x^2-2x(2q+1)+2q^2-3=0\leftrightarrow x=\frac{2(2q+1)\pm\sqrt{4(2q+1)^2-12(2q^2-3)}}{6}$$

The $|Delta$ must be greater then $0$ and so: $$q^2-2q-5<0 \leftrightarrow q<1-\sqrt{6} \lor q>1+\sqrt{6}$$