find the equation to the tangent to the curve y=2x^3+5y-4 at (-2,4)
Answers
Answer:
Answer
(a)
The equation of the given curve is y=x
2
−2x+7.
On differentiating with respect to x, we get:
dx
dy
=2x−2
The equation of the line is 2x−y+9=0.⇒y=2x+9
This is of the form y=mx+c.
Slope of the line =2
If a tangent is parallel to the line 2x−y+9=0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have: 2=2x−2
⇒2x=4⇒x=2
Now, at x=2
⇒y=2
2
−2×2+7=7
Thus, the equation of the tangent passing through (2,7) is given by,
y−7=2(x−2)
⇒y−2x−3=0
Hence, the equation of the tangent line to the given curve (which is parallel to line (2x−y+9=0) is y−2x−3=0.
(b)
The equation of the line is 5y−15x=13.
Slope of the line =3
If a tangent is perpendicular to the line 5y−15x=13,
then the slope of the tangent is
Slope of the line
−1
=
3
−1
.
⇒
dx
dy
=2x−2=
3
−1
⇒2x=
3
−1
+2
⇒2x=
3
5
⇒x=
6
5
Now, at x=
6
5
⇒y=
36
25
−
6
10
+7=
36
25−60+252
=
36
217
Thus, the equation of the tangent passing through (
6
5
,
36
217
) is given by,
(y−
36
217
)=−
3
1
(x−
6
5
)
⇒=
36
36y−217
−
18
1
(6x−5)
⇒36y−217=−2(6x−5)
⇒36y−217=−12x+10
⇒36y+12x−227=0
Hence, the equation of the tangent line to the given curve
(which is perpendicular to line 5y−15x=13) is 36y+12x−227=0