Question

The curve represented by $x = 3\left( {\cos t + \sin t} \right)$ and $y = 4\left( {\cos t - \sin t} \right)$ is
A. An ellipse
B. A parabola
C. A hyperbola
D. A circle

Answer 1

Hint: A curve is a smooth drawn figure or line in a plane that has a bend or turns in it. A circle, for example, is a curved shape. To solve the question square both the equations, add them and solve further and check which equation follows which curve.

Complete step by step solution: 
Given that,
$x = 3\left( {\cos t + \sin t} \right)$
$\dfrac{x}{3} = \cos t + \sin t - - - - - (1)$
And $y = 4\left( {\cos t - \sin t} \right)$
$\dfrac{y}{4} = \cos t - \sin t - - - - - (2)$
Adding the Square of equation (1) and (2)
${\left( {\dfrac{x}{3}} \right)^2} + {\left( {\dfrac{y}{4}} \right)^2} = {\left( {\cos t + \sin t} \right)^2} + {\left( {\cos t - \sin t} \right)^2}$
$\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = {\cos ^2}t + {\sin ^2}t + 2\sin t\cos t + {\cos ^2}t + {\sin ^2}t - 2\sin t\cos t$
$\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = 2\left( {{{\sin }^2}t + {{\cos }^2}t} \right)$
$\dfrac{{{x^2}}}{9} + \dfrac{{{y^2}}}{{16}} = 2$
$\dfrac{{{x^2}}}{{18}} + \dfrac{{{y^2}}}{{32}} = 1$
Required equation is an ellipse
Therefore, the correct option is (A)..

Note: The key concept involved in solving this problem is the good knowledge of Curve. Students must remember about the types of curves and their general equations. For Example. The general equation of the ellipse is $\dfrac{{{x^2}}}{a} + \dfrac{{{y^2}}}{b} = 1$.