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

Answer:

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

Step-by-step 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

$x=rcos\theta$

$y=rsin\theta$

$cos\theta=\frac{x}{r}$

$sin\theta=\frac{y}{r}$

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

$r=\frac{508y}{r}+\frac{41x}{r}$

$r^2=508y+41x$

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

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

To make completing square

$(x-\frac{41}{2})^2+(y-254)^2-\frac{1681}{4}-64516=0$

$(x-\frac{41}{2})^2+(y-254)^2+\frac{-1681-258064}{4}=0$

$(x-\frac{41}{2})^2+(y-254)^2-\frac{259745}{4}=0$

$(x-\frac{41}{2})^2+(y-254)^2=\frac{259745}{4}$

$(y-254)^2=\frac{259745}{4}-(x-\frac{41}{2})^2$

$y-254=\sqrt{\frac{259745}{4}-(x-\frac{41}{2})^2$

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