Question

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

Answer 1

Here, we will add both the parametric equation and then compare the obtained equation with the standard equation of the quadratic equation with two variables, $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ to find the value of $a$, $b$, $c$, $f$, $g$, and $h$. Then we will substitute the obtained values in the formula in the quadratic formula of the standard equation of two variables $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$, that is, $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$. If $\Delta \ne 0$, then substitute the values of $a$, $b$ and $h$ in the equation, ${h^2} = ab$.

Complete step-by-step answer:
We are given that the parametric equations.
$x = {t^2} + t + 1{\text{ ......eq.(1)}}$
$y = {t^2} - t + 1{\text{ ......eq.(2)}}$
Adding the equation $(1)$ with equation $(2)$, we get

$$\Rightarrow x + y = \left( {{t^2} + t + 1} \right) + \left( {{t^2} - t + 1} \right) \\\ \Rightarrow x + y = {t^2} + t + 1 + {t^2} - t + 1 \\\ \Rightarrow x + y = 2{t^2} + 2 \\\ \Rightarrow x + y = 2\left( {{t^2} + 1} \right){\text{ ......eq.(3)}} \\\$$

Subtracting the equation $(2)$ from equation $(1)$, we get

$$\Rightarrow x - y = \left( {{t^2} + t + 1} \right) - \left( {{t^2} - t + 1} \right) \\\ \Rightarrow x - y = {t^2} + t + 1 - {t^2} + t - 1 \\\ \Rightarrow x - y = 2t{\text{ ......eq.(4)}} \\\$$

Dividing the above equation by 2 on both sides, we get

$$\Rightarrow \dfrac{{x - y}}{2} = t \\\ \Rightarrow t = \dfrac{{x - y}}{2} \\\$$

Now, substituting the value of $t$ in the equation $(3)$, we get

$$\Rightarrow x + y = 2\left[ {1 + {{\left( {\dfrac{{x - y}}{2}} \right)}^2}} \right] \\\ \Rightarrow x + y = 2\left[ {\dfrac{{4 + {x^2} + {y^2} - 2xy}}{4}} \right] \\\ \Rightarrow x + y = \dfrac{{4 + {x^2} + {y^2} - 2xy}}{2} \\\$$

Multiplying the above equation by 2 on both sides, we get

$$\Rightarrow 2\left( {x + y} \right) = 2\left( {\dfrac{{4 + {x^2} + {y^2} - 2xy}}{2}} \right) \\\ \Rightarrow 2x + 2y = 4 + {x^2} + {y^2} - 2xy \\\$$

Subtracting the above equation by $2x + 2y$ on both sides, we get

$$\Rightarrow 2x + 2y - \left( {2x + 2y} \right) = 4 + {x^2} + {y^2} - 2xy - \left( {2x + 2y} \right) \\\ \Rightarrow 2x + 2y - 2x - 2y = 4 + {x^2} + {y^2} - 2xy - 2x - 2y \\\ \Rightarrow 0 = 4 + {x^2} + {y^2} - 2xy - 2x - 2y \\\ \Rightarrow 4 + {x^2} + {y^2} - 2xy - 2x - 2y = 0 \\\ \Rightarrow {x^2} + {y^2} - 2xy - 2x - 2y + 4 = 0{\text{ ......eq.(5)}} \\\$$

We will now compare the above equation with the standard equation of the quadratic equation with two variables, $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$ to find the value of $a$, $b$, $c$, $f$, $g$, and $h$.
$\Rightarrow a = 1$
$\Rightarrow b = 1$
$\Rightarrow c = 4$
$\Rightarrow f = - 1$
$\Rightarrow g = - 1$
$\Rightarrow h = - 1$
Substituting the above values in the formula in the quadratic formula of the standard equation of two variables $a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0$, that is, $\Delta = abc + 2fgh - a{f^2} - b{g^2} - c{h^2}$.

$$\Rightarrow \Delta = 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} \\\ \Rightarrow \Delta = 4 - 2 - 1 - 1 - 4 \\\ \Rightarrow \Delta = - 4 \\\$$

Therefore, $\Delta \ne 0$.
Substituting the values of $a$, $b$ and $h$ in the equation, ${h^2} = ab$.

$$\Rightarrow {\left( 1 \right)^2} = 1 \cdot 1 \\\ \Rightarrow 1 = 1 \\\$$

So, the given equation is an equation of a parabola.
Hence, option (c) is correct.

Note: In solving these types of questions, students should have to analyze the given parametric equations and try to eliminate the parameter using algebraic identities. Since the parameter can be eliminated by many methods as well. So it is advisable to analyze the equations to find the best possible method.