Eliminate the parameter and write the corresponding rectangular equation whose graph represents the curve.
x = secQ
Y = tanQ
Would I use cos^2Q + sin^2Q = 1?
x^2+y^2= 1/cos^2 + sin^2/cos^2
=x^2( 1+sin^2)=x^2(2-cos^2)=
= x^2 (2-1/x^2)= 2x^2-1
or
y^2-x^2=-1
x^2-y^2=1
check all that
Yes, you can use the trigonometric identity cos^2Q + sin^2Q = 1 to eliminate the parameter in this case.
Given:
x = secQ
y = tanQ
To eliminate the parameter Q and express the curve in terms of x and y, we can start by rewriting the secQ and tanQ in terms of cosQ and sinQ.
Recall that secQ is the reciprocal of cosQ and tanQ is the ratio of sinQ to cosQ:
x = secQ = 1/cosQ
y = tanQ = sinQ/cosQ
Now, let's rearrange the equations to solve for cosQ:
x = 1/cosQ
cosQ = 1/x
Similarly, solve y = sinQ/cosQ for sinQ:
y = sinQ/cosQ
sinQ = y * cosQ
Now, substitute the expressions for cosQ and sinQ back into the trigonometric identity:
(cosQ)^2 + (sinQ)^2 = 1
(1/x)^2 + (y * cosQ)^2 = 1
1/x^2 + y^2 * (cosQ)^2 = 1
Since we have x and y as the variables instead of Q, we obtain the rectangular equation:
1/x^2 + y^2 * (cosQ)^2 = 1
This equation represents the curve described by the parametric equations x = secQ and y = tanQ.