How do you convert $r = - 3 sec (θ)$ $r = -3sec(theta)$ to rectangular form?

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Answer 1 · 2016-03-22T13:02:50

$x = - 3$ $x=-3$ The graph is a vertical line.

$sec (θ) = \frac{1}{cos (θ)}$ $sec(theta) = 1/cos(theta)$

And cosine in $\frac{adjacent}{hypotenuse}$ $("adjacent")/("hypotenuse")$

So $sec (θ) = \frac{1}{cos (θ)} = \frac{hypotenuse}{adjacent}$ $sec(theta) =1/cos(theta)=("hypotenuse")/("adjacent")$

But hypotenuse $= \sqrt{x^{2} + y^{2}} and adjacent = x$ $= sqrt(x^2+y^2) " and adjacent "= x$

so $sec (θ) = \frac{\sqrt{x^{2} + y^{2}}}{x}$ $sec(theta)=sqrt(x^2+y^2)/x$

So we now have: $r = - 3 \frac{\sqrt{x^{2} + y^{2}}}{x}$ $" "r=-3sqrt(x^2+y^2)/x$

But $r = \sqrt{x^{2} + y^{2}}$ $" "r=sqrt(x^2+y^2)$ giving:

$\sqrt{x^{2} + y^{2}} = - 3 \frac{\sqrt{x^{2} + y^{2}}}{x}$ $" "sqrt(x^2+y^2) =-3sqrt(x^2+y^2)/x$

Divide both sides by $\sqrt{x^{2} + y^{2}}$ $sqrt(x^2+y^2)$ to get

$1 = - \frac{3}{x}$ $1=-3/x$

Multiply by $x$ $x$ to finish:

$x = - 3$ $x=-3$

Alternatively

Use $x = r cos θ$ $x=rcostheta$ and $sec θ = \frac{1}{cos θ}$ $sectheta = 1/costheta$

So we have

$r = - 3 sec θ$ $r=-3sectheta$

$r = \frac{- 3}{cos θ}$ $r = (-3)/costheta$

$r cos θ = - 3$ $rcostheta = -3$

$x = - 3$ $x=-3$

How do you convert r = -3sec(theta)r=−3sec(θ)r = -3sec(theta) to rectangular form?