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

$\textbf{Given:}$

$2y\,cos\theta=x\,sin\theta$

$2x\,sec\theta -y\,cosec\theta=3$

$\textbf{To find:}$

$\text{Relation between x and y}$

$\textbf{Solution:}$

$\text{The given equations can be written as}$

$2x\,sec\theta -y\,cosec\theta-3=0$

$x\,sec\theta-2y\,cosec\theta-0=0$

$\text{By cross multiplication rule, we get}$

$\dfrac{sec\theta}{0-6y}=\dfrac{cosec\theta}{-3x-0}=\dfrac{1}{-4xy+xy}$

$\dfrac{sec\theta}{-6y}=\dfrac{cosec\theta}{-3x}=\dfrac{1}{-3xy}$

$\implies$

$\dfrac{sec\theta}{-6y}=\dfrac{1}{-3xy}$

$sec\theta=\dfrac{-6y}{-3xy}$

$sec\theta=\dfrac{2}{x}$

$\text{Taking reciprocals, we get}$

$\bf\,cos\theta=\dfrac{x}{2}$

$\text{and}$

$\dfrac{cosec\theta}{-3x}=\dfrac{1}{-3xy}$

$cosec\theta=\dfrac{-3x}{-3xy}$

$cosec\theta=\dfrac{1}{y}$

$\text{Taking reciprocals, we get}$

$\bf\,sin\theta=\dfrac{y}{1}$

$\text{We know that,}$

$\bf\,cos^2\theta+sin^2\theta=1$

$(\frac{x}{2})^2+(\frac{y}{1})^2=1$

$\dfrac{x^2}{4}+\dfrac{y^2}{1}=1$

$\textbf{Answer:}$

$\text{The required relation is}$

$\boxed{\frac{x^2}{4}+\frac{y^2}{1}=1}$

Answer 2

Answer:

Step-by-step eAnswer: (1) 4

Solution:

Given,

2y cos θ = x sin θ….(i)

Squaring on both sides,

4y2 cos2θ = x2 sin2θ….(ii)

Again from the given,

2x sec θ – y cosec θ = 3

2x/cos θ – y/sin θ = 3

2x sin θ – y cos θ = 3 sin θ cos θ

2x sin θ – (x sin θ)/2 = 3 sin θ cos θ {from (i)}

3x sin θ = 6 sin θ cos θ

x = 2 cos θ….(iii)

From (ii) and (iii),

x2 + 4y2 = (2 cos θ)2 + [(x2 sin2θ)/cos2θ]

= 4 cos2θ + (4 cos2θ)sin2θ/cos2θ

= 4(cos2θ + sin2θ)

=4