Asked by Abbey
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 = secQ
Y = tanQ
Would I use cos^2Q + sin^2Q = 1?
Answers
Answered by
bobpursley
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
=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
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