Answers and Explanations 275
Answers
401–500
419.
y
x
= −
+
2
11
15
11
You know a point on the tangent line, so you just need to find the slope. Begin by
finding the derivative of 3(x
2
+ y
2
)
2
= 25(x
2
– y
2
):
3 2
2
2
25 2
2
2
2
x
y
x
y
dy
dx
x
y
dy
dx
+
+
=
−
(
)
Then enter the values x = 2 and y = 1 and solve for the slope
dy
dx
:
6 2 1 2 2 2 1
2 5 2 2 2 1
6 5 4 2
2
2
+
+
=
−
+
(
)
( ) ( )
( ) ( )
( )
dy
dx
dy
dx
dy
d dx
dy
dx
dy
dx
dy
dx
=
−
= −
= −
25 4 2
110
2 0
2
11
The tangent line has a slope of − 2
11
and passes through (2, 1), so its equation is
y
x
y
x
− = −
−
= −
+
1
2
11
2
2
11
15
11
(
)
Answers
401–500
419.
y
x
= −
+
2
11
15
11
You know a point on the tangent line, so you just need to find the slope. Begin by
finding the derivative of 3(x
2
+ y
2
)
2
= 25(x
2
– y
2
):
3 2
2
2
25 2
2
2
2
x
y
x
y
dy
dx
x
y
dy
dx
+
+
=
−
(
)
Then enter the values x = 2 and y = 1 and solve for the slope
dy
dx
:
6 2 1 2 2 2 1
2 5 2 2 2 1
6 5 4 2
2
2
+
+
=
−
+
(
)
( ) ( )
( ) ( )
( )
dy
dx
dy
dx
dy
d dx
dy
dx
dy
dx
dy
dx
=
−
= −
= −
25 4 2
110
2 0
2
11
The tangent line has a slope of − 2
11
and passes through (2, 1), so its equation is
y
x
y
x
− = −
−
= −
+
1
2
11
2
2
11
15
11
(
)