Eliminating parameter from the given equations

Question

The curve described parametrically by $x=t^2+t+1$ and $y=t^2-t+1$ represents?

My attempt:

No clue on this one. I am not able to eliminate t from these equations. However I do know that the equation represented is a parabola...

From this, https://ggbm.at/mXbwME9P ( Click on the play button at the bottom right of the slider )

Can anyone help?

Answer 1

$x+y=2(t^2+1)$ and $(x-y)=2t$

Now eliminating $t$ from these equations

$$\frac{x+y}{2}=2\bigg[\bigg(\frac{x-y}{2}\bigg)^2+1\bigg]$$

So $$\frac{x+y}{2}=\frac{(x-y)^2}{2}+2 $$

Put $x+y=X$ and $x-y=Y$. Then equation convert into $$X=Y^2+4$$

Which Represent Parabola.

Answer 2

$$x-y=2t\iff t=?$$

$$x+y=2(t^2+1)\iff t^2=?$$

Use $t^2=(t)^2$ to eliminate $t$

Use Rotation of conic sections to identify the conic.

Answer 3

One has $x-y=2t$. Therefore $$4x=4t^2+4t+4=(x-y)^2+2(x-y)+4$$ etc.

Answer 4

You can use the method of resultants to eliminate $t$. The polynomials $t^2+t+1-x$ and $t^2-t+1-y$ have common roots when the determinant of the Sylvester matrix $$\begin{bmatrix}t^2&t&1-x&0 \\ 0&t^2&t&1-x \\ t^2&-t&1-y&0 \\ 0&t^2&-t&1-y\end{bmatrix}$$ vanishes. Expanding this determinant and setting it equal to zero results in the equation $$t^4(x^2-2xy+y^2-2x-2y+4) = 0.$$ We want this to vanish for all values of $t$, so we must have $$x^2-2xy+y^2-2x-2y+4 = 0.$$ The discriminant of this equation is equal to zero, so it represents a parabola.