If 2y cos θ = x sin θ and 2x sec θ - y cosec θ = 3, then x2 + 4y2 =

Question

If 2y cos θ = x sin θ and 2x sec θ - y cosec θ = 3, then x2 + 4y2 = (A) 4 (B) -4 (C) ±4 (D) None of these

Tardigrade Admin · Accepted Answer

Given that 2y cos θ = x sin θ ...(i) and 2x sec θ - y cosec θ = 3 ...(ii) ⇒ (2x/cos θ ) - (y/sin θ ) = 3 ⇒ 2x sin θ - y cos θ -3sin θ cos θ = 0 ...(iii) Solving (i) and (iii), we get y = sin θ and x = 2 cos θ Now, x2 + 4y2 = 4cos2 θ + 4sin2 θ = 4(cos2 θ + sin2 θ ) = 4

Q. If 2ycosθ=xsinθand2xsecθ−ycosecθ=3, then x2+4y2=

4

-4

±4

None of these

Given that 2ycosθ=xsinθ...(i) and 2xsecθ−ycosecθ=3...(ii) ⇒cosθ2x​−sinθy​=3 ⇒2xsinθ−ycosθ−3sinθcosθ=0...(iii) Solving (i) and (iii), we get y=sinθ and x=2cosθ Now, x2+4y2=4cos2θ+4sin2θ =4(cos2θ+sin2θ)=4

Q. If $2 y cos θ = x s in θ an d 2 x sec θ - y cosec θ = 3,$ then $x^{2} + 4 y^{2} =$