Question

A curve with polar equation r=508sinθ+41cosθ r=508sin⁡θ+41cos⁡θ represents a line. This line has a Cartesian equation of the form y=mx+by=mx+b ,where mm and bb are constants. Give the formula for yy in terms of xx. For example, if the line had equation y=2x+3y=2x+3 then the answer would be 2∗x+32∗x+3 .

Answer 1

`y=\sqrt{(259745)/(4)-(x-(41)/(2))^2}+254`

Explanation:

We are given that a curve with polar equation

`r=508sin\theta+41 cos\theta`

We have to find the equation of curve in Cartesian form.

We know that

`x=rcos\theta`

`y=rsin\theta`

`cos\theta=(x)/(r)`

`sin\theta=(y)/(r)`

Squaring both sides and then adding

`x^2+y^2=r^2(cos^2\theta+sin^2\theta)=r^2`

Because
`cos^2\theta+sin^2\theta=1`

Substitute the values then we get

`r=(508y)/(r)+(41x)/(r)`

`r^2=508y+41x`

`x^2+y^2=508y+41x`

`x^2-41x+y^2-508y=0`

To make completing square

`(x-(41)/(2))^2+(y-254)^2-(1681)/(4)-64516=0`

`(x-(41)/(2))^2+(y-254)^2+(-1681-258064)/(4)=0`

`(x-(41)/(2))^2+(y-254)^2-(259745)/(4)=0`

`(x-(41)/(2))^2+(y-254)^2=(259745)/(4)`

`(y-254)^2=(259745)/(4)-(x-(41)/(2))^2`

`y-254=\sqrt{(259745)/(4)-(x-(41)/(2))^2`

`y=\sqrt{(259745)/(4)-(x-(41)/(2))^2}+254`