find an equation in rectangular coordinates p+3=e and r-2 = 3cosΘ

Question

alenahelenash · Accepted Answer

We can start by expressing the equation

r - 2 = 3 \cos θ

in rectangular coordinates. Recall that

r^{2} = x^{2} + y^{2}

, so we can square both sides of the equation and substitute

x^{2} + y^{2}

for

r^{2}

:

(x^{2} + y^{2}) - 4 = 9 \cos^{2} θ

Using the identity

\cos^{2} θ + \sin^{2} θ = 1

, we can substitute

1 - \sin^{2} θ

for

\cos^{2} θ

:

(x^{2} + y^{2}) - 4 = 9 (1 - \sin^{2} θ)

Simplifying:

x^{2} + y^{2} = 13 + 9 \sin^{2} θ

Now, we need to find an equation for

p + 3 = e

in rectangular coordinates. Recall that

p = \sqrt{x^{2} + y^{2}}

and

e = \frac{x}{p}

, so we have:

\sqrt{x^{2} + y^{2}} + 3 = \frac{x}{\sqrt{x^{2} + y^{2}}}

Squaring both sides and multiplying by

x^{2} + y^{2}

, we get:

(x^{2} + y^{2})^{2} + 6 (x^{2} + y^{2}) + 9 = x^{2}

Simplifying:

x^{4} + 2 x^{2} y^{2} + y^{4} + 6 x^{2} + 6 y^{2} + 9 = x^{2}

Moving all the terms to one side, we get:

x^{4} + 2 x^{2} y^{2} + y^{4} + 5 x^{2} - x^{2} + 6 y^{2} + 9 = 0

Simplifying:

x^{4} + 2 x^{2} y^{2} + y^{4} + 4 x^{2} + 6 y^{2} + 9 = 0

Therefore, the equation in rectangular coordinates is:

x^{4} + 2 x^{2} y^{2} + y^{4} + 4 x^{2} + 6 y^{2} + 9 = 0

find an equation in rectangular coordinates p+3=e and