(1) Eliminate theta (ii) x=2 cos theta-3sin theta, y=cos theta+ 2 sin theta
Answers
Step-by-step explanation:
Given x=asinθ⟹
a
x
=sinθ
y=acosθ⟹
a
y
=cosθ
As we know that
sin
2
θ+cos
2
θ=1
(
a
x
)
2
+(
a
y
)
2
=1
⟹x
2
+y
2
=a
2
Answer:
Given that,
x = 2 cosθ - 3 sinθ ...(i)
y = cosθ + 2 sinθ
⇒ 2y = 2 cosθ + 4 sinθ ...(ii)
Now, (ii) - (i) ⇒
2y - x = 2 cosθ + 4 sinθ - 2 cosθ + 3 sinθ
⇒ 2y - x = 7 sinθ ...(iii)
Again, (ii) × 7 ⇒
7y = 7 cosθ + 2 (7 sinθ)
⇒ 7y = 7 cosθ + 2 (2y - x), by (iii)
⇒ 7 cosθ = 3y + 2x ...(iv)
Finally, we have
cosθ = \frac{\text{3y+2x}}{7}
7
3y+2x
sinθ = \frac{\text{2y-x}}{7}
7
2y-x
We know that,
sin²θ + cos²θ = 1
\implies {(\frac{2y - x}{7})}^{2} + {(\frac{3y + 2x}{7}})^{2} = 1⟹(
7
2y−x
)
2
+(
7
3y+2x
)
2
=1
\implies \frac{4{y}^{2} - 4xy + {x}^{2}}{49} + \frac{9{y}^{2} + 12xy + 4{x}^{2}}{49} = 1⟹
49
4y
2
−4xy+x
2
+
49
9y
2
+12xy+4x
2
=1
\implies \frac{4{y}^{2} - 4xy + {x}^{2} + 9{y}^{2} + 12xy + 4{x}^{2}}{49}=1⟹
49
4y
2
−4xy+x
2
+9y
2
+12xy+4x
2
=1
⇒ 5x² + 13y² + 8xy = 49,
which is the required equation