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Mathematics
Problem
For the curve defined parametrically by
x
=
3
t
2
+
1
x=3t^{2}+1
x
=
3
t
2
+
1
and
y
=
2
t
−
5
y=2t-5
y
=
2
t
−
5
, find the equation of the tangent at the point corresponding to
t
=
2
t=2
t
=
2
. a)
6
x
−
y
−
19
=
0
6x-y-19=0
6
x
−
y
−
19
=
0
b)
x
+
6
y
−
19
=
0
x+6y-19=0
x
+
6
y
−
19
=
0
c)
x
−
6
y
−
19
=
0
x-6y-19=0
x
−
6
y
−
19
=
0
d)
x
−
6
y
+
19
=
0
x-6y+19=0
x
−
6
y
+
19
=
0
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