Question

The curve represented by $x=3(\cos t+\sin t),y=4(\cos t-\sin t)$is
(a) An ellipse
(b) A Circle
(c) A hyperbola
(d) A parabola

Answer 1

Squaring both the given equations and then add them. Cancel the like terms. Now compare the result obtained with predefined equations of various geometrical shapes. Use the formula of trigonometric ratios to simplify the equation.

Complete step-by-step answer:
The given equations are,
$x=3(\cos t+\sin t)$
It can be written as
$\dfrac{x}{3}=(\cos t+\sin t)$
Squaring both the sides, we get
${{\left( \dfrac{x}{3} \right)}^{2}}={{(\cos t+\sin t)}^{2}}$
Now we know the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , using this in above expression, we get
$\Rightarrow \dfrac{{{x}^{2}}}{{{3}^{2}}}={{\cos }^{2}}t+{{\sin }^{2}}t+2\sin t.\cos t$
But we know,$\left[ {{\cos }^{2}}t+{{\sin }^{2}}t=1 \right]$, so the above equation becomes,
$\dfrac{{{x}^{2}}}{9}=1+2\sin t.\cos t..........(i)$
Now consider the other equation,
$y=4(\cos t-\sin t)$
It can be written as
$\dfrac{y}{4}=(\cos t-\sin t)$
Squaring both the sides, we get
${{\left( \dfrac{y}{4} \right)}^{2}}={{(\cos t-\sin t)}^{2}}$
Now we know the formula, ${{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$ , using this in above expression, we get
$\Rightarrow \dfrac{{{y}^{2}}}{{{4}^{2}}}={{\cos }^{2}}t+{{\sin }^{2}}t-2\sin t.\cos t$
But we know $\left[ {{\cos }^{2}}t+{{\sin }^{2}}t=1 \right]$, so the above equation becomes,
$\dfrac{{{y}^{2}}}{16}=1-2\sin t.\cos t......(ii)$
Adding equation (i) and (ii) and cancelling the like terms, we get

& \text{ }\dfrac{{{x}^{2}}}{9}=1+2\sin t\cos t \\\ & (+)\dfrac{{{y}^{2}}}{16}=1-2\sin t\cos t \\\ \end{aligned}}{\dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{16}=1+1}$$ $\Rightarrow \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{16}=2$ Dividing throughout by ‘2’, we get $$\Rightarrow \dfrac{1}{2}\left( \dfrac{{{x}^{2}}}{9}+\dfrac{{{y}^{2}}}{16} \right)=1$$ Now, ‘18’ can be written as ${{\left( 3\sqrt{2} \right)}^{2}}$ , so the above equation becomes, $\Rightarrow \dfrac{{{x}^{2}}}{{{(3\sqrt{2})}^{2}}}+\dfrac{{{y}^{2}}}{32}=1$ Similarly, ‘32’ can be written as ${{\left( 4\sqrt{2} \right)}^{2}}$ , so the above equation becomes, $\Rightarrow \dfrac{{{x}^{2}}}{{{(3\sqrt{2})}^{2}}}+\dfrac{{{y}^{2}}}{{{(4\sqrt{2})}^{2}}}=1$ Comparing this with general equation of ellipse, that is $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ So we see that the given set of equations represents an ellipse. **So, the correct answer is “Option (a)”.** **Note:** Whenever this type of question is given, that involves sine and cosine functions, squaring is must and then add the result. This is the easiest way to solve it. Suppose that we consider the operation $\dfrac{x}{y}$ , we get a different result.