Find the equation of all lines having slope 2 and being tangent to the curve y+2/x-3=0
Answers
Answered by
1
Answer:
0
Step-by-step explanation:
The curve is y+
x−3
2
=0
Slope of the tangent to the curve at point (x,y) is
dx
dy
dx
dy
+
dx
d
(
x−3
2
)=0
⇒
dx
dy
=−
dx
d
(
x−3
2
)
⇒
dx
dy
=−(
(x−3)
2
−2
)
Given that slope=2
hence,
dx
dy
=2
⇒(
(x−3)
2
2
)=2
⇒(x−3)
2
=1
⇒x−3=±1
∴x−3=1,x−3=−1
∴x=4,2
If x=2⇒y=
x−3
−2
=
2−3
−2
=2
Thus, the point is (2,2)
If x=4⇒y=
x−3
−2
=
4−3
−2
=−2
Thus, the point is (4,−2)
Thus, there are two tangents to the curve with slope 2 and passing through points (2,2) and (4,−2)
Equation of tangent through (2,2) is
y−2=2(x−2)
y−2x+2=0
Equation of tangent through (4,−2) is
y−(−2)=2(x−4)
y−2x+10=0
Answered by
0
Answer:
-2
Step-by-step explanation:
y+2/x-3=0
y+2=0
y=0-2
y=-2
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