Respuesta :

[tex]x+y=k\implies y=k-x[/tex]

Substitute this into the parabolic equation,

[tex]k-x=6-(x-3)^2\implies x^2-7x+(k+3)=0[/tex]

We're told the line [tex]x+y=k[/tex] intersects [tex]C[/tex] twice, which means the quadratic above has two distinct real solutions. Its discriminant must then be positive, so we know

[tex](-7)^2-4(k+3)=49-4(k+3)>0\implies k<\dfrac{37}4=9.25[/tex]

We can tell from the quadratic equation that [tex]C[/tex] has its vertex at the point (3, 6). Also, note that

[tex]-1\le\sin t\le1\implies3\le3+2\sin t\le5\implies3\le x\le5[/tex]

and

[tex]-1\le\cos2t\le1\implies2\le4+2\cos2t\le6\implies2\le y\le6[/tex]

so the furthest to the right that [tex]C[/tex] extends is the point (5, 2). The line [tex]x+y=k[/tex] passes through this point for [tex]2+5=k\implies k=7[/tex]. For any value of [tex]k<7[/tex], the line [tex]x+y=k[/tex] passes through [tex]C[/tex] either only once, or not at all.

So [tex]7\le k<9.25[/tex]; in set notation,

[tex]\{k\mid 7\le k<9.25\}[/tex]

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