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# The curve described parametrically by $$x = {t^2} + t + 1,y = {t^2} - t + 1$$ representsA) A pair of straight lineB) An eclipseC) A parabolaD) A hyperbola

Last updated date: 16th Jun 2024
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Hint: We have given two equations in the form of x and y, compare both equations and add and subtract it, after that compare both the equations, it will be similar to the parabola equation, we will get an answer.

We have, $$x = {t^2} + t + 1$$ .... (i)
and $$y = {t^2} - t + 1$$ .... (ii)
Now, $$x + y = 2(1 + {t^2})$$ .... (iii)
and $$x - y = 2t$$.... (iv)
Now, from Eqs. (iii) and (iv), we get
\eqalign{ & x + y = 2[1 + {\left( {\dfrac{{(x - y)}}{2}} \right)^2}] \cr & \Rightarrow x + y = 2[\dfrac{{4 + {x^2} + {y^2} - 2xy}}{4}] \cr}
$$\Rightarrow {x^2} + {y^2} - 2xy - 2x - 2y + 4 = 0\;{\text{ }}\;{\text{ }}\;$$ .... (v)
On comparing with, we get
$$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$
We get, $$a = 1,b = 1,c = 4,h = - 1,g = - 1,f = - 1$$
$$\vartriangle = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$$
Now,
\eqalign{ & \vartriangle = 1 \cdot 1 \cdot 4 + 2\left( { - 1} \right)\left( { - 1} \right)\left( { - 1} \right) - 1 \times {\left( { - 1} \right)^2} - 1 \times {\left( { - 1} \right)^2} - 4{\left( { - 1} \right)^2} \cr & = 4 - 2 - 1 - 1 - 4 \cr & = - 4 \cr}
, therefore, $$\vartriangle \ne 0$$
and $$ab - {h^2} = 1 \cdot 1 - {\left( 1 \right)^2} = 1 - 1 = 0$$
So, it is the equation of a parabola.

Note: We knew the equation of parabola, i.e. $$a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$$. After comparing both the given equations it gets similar to the parabola equation, so the answer is the equation is of parabola.