Question

If 2y cos theta=x sin theta and 2x sec theta-y cosec theta=3,then the relation between x and y is

Answer 1

2y/x=sin theta/cos theta i think

Answer 2

Let cos theta be u and sin theta be v
2yu = xv --(1)
(2x/u) - (y/v) =3 --(2) [because sec theta = (1/cos theta) and cos theta = (1/sin theta)

From (2), 2xv - yu = 3uv
2xv - 3uv = yu
y = (2xv - 3vu) / u --(3)

Sub (3) into (1),
2u[(2xv-3uv)/u] = xv
*cancelling u from numerator and denominator*
2(2xv-3uv) = xv
4xv-6uv = xv
3xv = 6uv
*cancelling v from both sides*
3x = 6u
x = 2u --(4)

Sub (4) into (3)
y = [2(2u)v - 3vu)] / u
y = (4vu -3vu) / u
y = vu / u
y = v --(5)

From (4), x = 2u
x = 2cos theta and y = sin theta

Answer 3

In this case we use the method of substitution to simplify the equations to solve this problem, hope it helps! :)

Answer 4

nope .i need the equation ................@HatcrewS .something went wrong here.

Answer 5

@ParthKohli

Answer 6

@uskidoll

Answer 7

@janani1

@HatcrewS is correct, you just need to continue where she left off to get it in terms of only"x" and "y"
x = 2 cos ---> x/2 = cos
y = sin

\[\sin^2 \theta + \cos^2 \theta = 1\]
\[\rightarrow y^2 + (\frac{x}{2})^2 = 1\]
\[y^2 + \frac{x^2}{4} = 1\]

this is the relation between x and y ---> an ellipse with major axis of 4 and minor axis of 2

Answer 8

nice one @HatcrewS