Question

The curve described parametrically by $$x = {t^2} + t + 1,y = {t^2} - t + 1$$ represents
A) A pair of straight line
B) An eclipse
C) A parabola
D) A hyperbola

Answer 1

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.

Complete step-by-step 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.