Expert Answer to "A polar conic section Consider the equation r2=sec2θ...."

pedzenekO

Answered question

2020-12-30

A polar conic section Consider the equation

r^{2} = s e c 2 θ

. Convert the equation to Cartesian coordinates and identify the curve

Answer & Explanation

Cullen

Skilled2020-12-31Added 89 answers

Given: we have an equation in polar form $r^{2} = \sec 2 θ$ Part a) to determinate the equation to Cartesian coordinates and identify the curve Explain: to convert the equation we can write the equation sa follows $r = \sqrt{\sec 2 θ}$
$r = 1 \sqrt{\cos 2 θ}$ We know these relations $x = r \cos θ, y = r \sin θ, r = \sqrt{x^{2} + y^{2}}$ Can be written $\cos θ$ $= x / r, \sin θ = y / r$ $r = \frac{1}{\sqrt{\cos^{2} θ}} = \frac{1}{\sqrt{\cos^{2} θ - \sin^{2} θ}} [\cos^{2} θ = \cos^{2} θ - \sin^{2} θ]$

Putting in the equation we have $r = {\frac{1}{\sqrt{\frac{x}{r}}}}^{2} - {\frac{y}{r}}^{2}$ $\frac{x^{2}}{r^{2}} - \frac{y^{2}}{r^{2}} = \frac{1}{r^{2}}$ We have the equation $x^{2} - y^{2} = 1$

Do you have a similar question?

Recalculate according to your conditions!

New Questions in High school geometry

Didn't find what you were looking for?