Question

The curve described para metrically by $x=t^2+2t-1$, $y=3t+5$ represents

• A. an ellipse
• B. a hyperbola
• C. a parabola
• D. a circle

Answer 1

The correct option is C

a parabola

Explanation for correct option:

Identifying the type of curve:

Given that

$x=t^2-t-1\dots \dots \left(1\right)$

$y=3t+5\dots \dots \left(2\right)$

Solving equation $\left(1\right)$ we get,

$$\begin{gathered} x=t^2+2t-1 \\ =t^2+2t+1-2 \\ \Rightarrow x={\left(t+1\right)}^2-2 \\ \Rightarrow x+2={\left(t+1\right)}^2 \\ \Rightarrow t+1={\left(x+2\right)}^{\frac{1}{2}}\dots \dots \left(3\right) \end{gathered}$$

Solving equation $\left(2\right)$ we get,

$$\begin{gathered} y=3t+5 \\ =3t+3+2 \\ \Rightarrow y=3\left(t+1\right)+2 \\ \Rightarrow y-2=3\left(t+1\right)\dots \dots \left(4\right) \end{gathered}$$

Substituting $t+1={\left(x+2\right)}^{\frac{1}{2}}$ in equation $\left(4\right)$

$$\begin{gathered} y-2=3{\left(x+2\right)}^{\frac{1}{2}} \\ y-2=3{\sqrt{x+2}}^{} \end{gathered}$$

Squaring on both sides we get

$$\begin{gathered} \Rightarrow {\left(y-2\right)}^2={\left(3{\sqrt{x+2}}^{}\right)}^2 \\ \Rightarrow {\left(y-2\right)}^2=9\left(x+2\right) \end{gathered}$$

Let $y-2=Y$ and $X=x+2$

We get,

$\Rightarrow Y^2=9X$

The equation is of the form of $Y^2=CX$ which the general form of a parabola.

Hence, option (C) is the correct answer