Question

The curve represented by$x=3(\cos t+\sin t)$and $y=4(\cos t-\sin t)$ is

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

Answer 1

The correct option is A

an ellipse

Explanation for correct option:

Identifying the type of curve:

Given that the equation of two curve is

$x=3(\cos t+\sin t)$

$\Rightarrow \frac{x}{3}=\cos t+\sin t\dots \dots \left(1\right)$

And

$y=4(\cos t-\sin t)$

$\Rightarrow \frac{y}{4}=\cos t-\sin t\dots \dots \left(2\right)$

Now, squaring equation $\left(1\right)$

$$\begin{gathered} {\left(\frac{x}{3}\right)}^2={\left(\cos t+\sin t\right)}^2 \\ \frac{x^2}{9}={\left(\cos t+\sin t\right)}^2\dots \dots \left(3\right) \end{gathered}$$

Squaring equation $\left(2\right)$

$$\begin{gathered} {\left(\frac{y}{4}\right)}^2={\left(\cos t-\sin t\right)}^2 \\ {\frac{y}{16}}^2={\left(\cos t-\sin t\right)}^2\dots \dots \left(4\right) \end{gathered}$$

Adding equations $\left(3\right)$and $\left(4\right)$ we get

$\begin{array}{rcl}\frac{x^2}{9}+{\frac{y}{16}}^2& =& {\left(\cos t+\sin t\right)}^2+{\left(\cos t-\sin t\right)}^2\\ & =& 2\left({\cos }^{2}t+{\sin }^{2}t\right)\\ \frac{x^2}{9}+{\frac{y}{16}}^2& =& 2\left[\because {\cos }^{2}t+{\sin }^{2}t=1\right]\end{array}$

Therefore,$\frac{x^2}{18}+{\frac{y}{32}}^2=1$ is in the form of $\frac{x^2}{a^2}+{\frac{y}{b^2}}^2=1$ which is an ellipse.

Hence, option (A) is the correct answer.