Question

The curve represented by x = 2 (cos t + sin t),
y = 5 (cos t - sin t) is
(a) a circle
(b) a parabola
(c) an ellipse
(d) a hyperbola​

Answer 1

Answer:

x/2 = sin t + cos t

y/5 = cos t - sin t

Squaring both the relations and then adding to each other,

(x/2)^2 + (y/5)^2 = 2 (sin^2 t + cos^2 t) + 2 sint cost - 2 sint cost

x^2/4 + y^2/25 = 2

or,

x^2/8+y^2/50=1

So, it's an ellipse !! ｡◕‿◕｡

Answer 2

Answer:

$\large\bold\red{(c)\:an\:ellipse}$

Step-by-step explanation:

Given a curve represented by,

$x = 2( \cos(t) + \sin(t) ) \\ \\ = > \frac{x}{2} = \cos(t) + \sin(t) \: \: \: \: \: ..........(1)$

And,

$y = 5( \cos(t ) - \sin(t) ) \\ \\ = > \frac{y}{5} = \cos(t ) - \sin(t) \: \: \: \: \: ...........(2)$

Now,

Squaring and adding both the Equation,

We get,

$= > {( \frac{x}{2}) }^{2} + {( \frac{y}{5} )}^{2} = {( \cos t + \sin t) }^{2} + {( \cos t - \sin t) }^{2} \\ \\ = > \frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{25} = { \cos}^{2} t + { \sin}^{2} t + \cancel{ 2 \cos(t) \sin(t) }+ { \cos }^{2} t + { \sin }^{2}t - \cancel{2 \cos(t) \sin(t) } \\ \\ = > \frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{25} = 2( { \cos }^{2} t + { \sin }^{2} t) \\ \\ = > \frac{ {x}^{2} }{4} + \frac{ {y}^{2} }{25} = 2 \\ \\ = > \frac{1}{2} \times ( \frac{ {x}^{2} }{ {2}^{2} } + \frac{ {y}^{2} }{ {5}^{2} } ) = \frac{1}{2} \times 2 \\ \\ = > \frac{ {x}^{2} }{8} + \frac{ {y}^{2} }{50} = 1 \\ \\ = >\large\bold{ \frac{ {x}^{2} }{ {(2 \sqrt{2}) }^{2} } + \frac{ {y}^{2} }{ {(5 \sqrt{2} )}^{2} } = 1}$

Clearly,

It's in the form of Equation of an ellipse, i.e,

$\large \boxed{ \bold \purple{\frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } = 1}}$

Hence,

It represents (c) an ellipse.