Animated Solution for Mathematics - Conic Sections: The curve described parametrically by $x=t^2+t+1,y=t^2-t+1$ represents

Have you ever looked at a set of parametric equations and felt like you were staring at a secret code? You are given $x=t^2+t+1$ and $y=t^2-t+1$.

At first glance, they look like two separate, unrelated paths. But in the world of coordinate geometry, these are just two different ways of describing the same object.

Our mission today is to act as mathematical detectives and decode this parametric mystery into a familiar Cartesian form.

The Algebraic Dance

To uncover the true identity of this curve, we must eliminate the parameter $t$. Think of $t$ as a bridge between $x$ and $y$.

If we can remove the bridge, we can see the landscape of the curve directly in the $xy$-plane. Look at the structure of our equations; notice how the $t^2$ and the constant $1$ are identical in both. This is a huge hint!

If we subtract the second equation from the first, we get:

This is beautiful! The $t^2$ and the $1$ vanish, leaving us with $2t=x-y$, or simply $t=\frac{x-y}{2}$. We have successfully captured $t$ in terms of $x$ and $y$.

Now, what if we add them? Adding the equations gives us:

We can factor out the $2$ to get $x+y=2(t^2+1)$. Now we have an expression for $t^2$ as well. The bridge is crumbling, and the Cartesian form is emerging!

The Transformation

Now, let's perform the substitution. We take our expression for $t$ and plug it into our sum equation:

Don't let the brackets scare you. Let's expand this step-by-step. Squaring the fraction gives us $\frac{(x-y{)}^2}{4}$. Multiplying by $2$ yields:

To make this look like a standard equation, let's multiply everything by $2$ to clear the denominator:

Expanding the perfect square $(x-y{)}^2=x^2-2xy+y^2$, we get:

Finally, moving everything to one side, we arrive at the general second-degree equation:

The Classification

We have arrived at the final stage of our journey. We have a general second-degree equation, and we need to identify its shape.

We compare our equation to the standard form $a{x}^2+2hxy+b{y}^2+2gx+2fy+c=0$. By matching the coefficients, we find $a=1$, $b=1$, $c=4$, $h=-1$, $g=-1$, and $f=-1$.

First, we calculate the discriminant $\Delta$ to ensure it is a non-degenerate conic:

Substituting our values, we get $\Delta =-4$. Since $\Delta eq 0$, we are dealing with a proper conic section.

Now, for the final test: the relationship between $h^2$ and $ab$. We calculate $h^2=(-1{)}^2=1$ and $ab=(1)(1)=1$.

Since $h^2=ab$, the curve is a Parabola.

We started with two parametric equations that seemed disconnected, and through the power of algebraic manipulation, we revealed a beautiful parabola.

This is the essence of JEE mathematics—taking a complex, hidden form and using your toolkit to reveal the elegant, simple geometry underneath. Keep practicing, keep questioning, and keep falling in love with the process!