The curve represented by $x = 3 (c o s t + s i n t), y = 4 (c o s t - s i n t)$ is

Ellipse

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Parabola

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Hyperbola

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Circle

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Solution

The correct option is A Ellipse
Given $\frac{x}{3} = cos t + sin t & \frac{y}{4} = cos t - sin t$

Squaring these two,

$\Rightarrow \frac{x^{2}}{9} = 1 + 2 sin t . cos t . . . (i)$

$\frac{y^{2}}{16} = 1 - 2 sin t cos t . . . . (i i)$
Adding (i) & (ii)

$\frac{x^{2}}{9} + \frac{y^{2}}{16} = 2 \Rightarrow \frac{x^{2}}{18} + \frac{y^{2}}{32} = 1$

So the equation represents ellipse.

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Similar questions

x = 3 (cos t + sin t)

y = 4 (cos t - sin t)

x = 2 (cos t - sin t)

y = 3 (c o s t + s i n t)

x = 3 cos t + 3 sin t

y = 4 cos t - 4 sin t

x = 3 (cos t + sin t); y = 2 (cos t - sin t)

x = sin (t + \frac{7 π}{12}) + sin (t - \frac{π}{12}) + sin (t + \frac{3 π}{12}),

y = cos (t + \frac{7 π}{12}) + cos (t - \frac{π}{12}) + cos (t + \frac{3 π}{12})

\frac{d}{d t} (\frac{x}{y} - \frac{y}{x})

t = \frac{π}{8}

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