Answer:
[tex]y=\sqrt{\frac{259745}{4}-(x-\frac{41}{2})^2}+254[/tex]
Step-by-step explanation:
We are given that a curve with polar equation
[tex]r=508sin\theta+41 cos\theta[/tex]
We have to find the equation of curve in Cartesian form.
We know that
[tex]x=rcos\theta[/tex]
[tex]y=rsin\theta[/tex]
[tex]cos\theta=\frac{x}{r}[/tex]
[tex]sin\theta=\frac{y}{r}[/tex]
Squaring both sides and then adding
[tex]x^2+y^2=r^2(cos^2\theta+sin^2\theta)=r^2[/tex]
Because [tex]cos^2\theta+sin^2\theta=1[/tex]
Substitute the values then we get
[tex]r=\frac{508y}{r}+\frac{41x}{r}[/tex]
[tex]r^2=508y+41x[/tex]
[tex]x^2+y^2=508y+41x[/tex]
[tex]x^2-41x+y^2-508y=0[/tex]
To make completing square
[tex](x-\frac{41}{2})^2+(y-254)^2-\frac{1681}{4}-64516=0[/tex]
[tex](x-\frac{41}{2})^2+(y-254)^2+\frac{-1681-258064}{4}=0[/tex]
[tex](x-\frac{41}{2})^2+(y-254)^2-\frac{259745}{4}=0[/tex]
[tex](x-\frac{41}{2})^2+(y-254)^2=\frac{259745}{4}[/tex]
[tex](y-254)^2=\frac{259745}{4}-(x-\frac{41}{2})^2[/tex]
[tex]y-254=\sqrt{\frac{259745}{4}-(x-\frac{41}{2})^2[/tex]
[tex]y=\sqrt{\frac{259745}{4}-(x-\frac{41}{2})^2}+254[/tex]
