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Question

Find the equation of the tangent line to the curve y=x22x+7 which is.
(a) parallel to the line 2xy+9=0.
(b) perpendicular to the line 5y15x=13.

Solution
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(a)
The equation of the given curve is y=x22x+7.
On differentiating with respect to x, we get:
dydx=2x2
The equation of the line is 2xy+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 2xy+9=0, then the slope of the tangent is equal to the slope of the line.
Therefore, we have: 2=2x2
2x=4x=2
Now, at x=2
y=222×2+7=7
Thus, the equation of the tangent passing through (2,7) is given by,
y7=2(x2)
y2x3=0
Hence, the equation of the tangent line to the given curve (which is parallel to line (2xy+9=0) is y2x3=0.

(b)
The equation of the line is 5y15x=13.
Slope of the line =3
If a tangent is perpendicular to the line 5y15x=13,
then the slope of the tangent is 1Slope of the line=13.
dydx=2x2=13
2x=13+2
2x=53
x=56
Now, at x=56
y=2536106+7=2560+25236=21736
Thus, the equation of the tangent passing through (56,21736) is given by,
(y21736)=13(x56)
=36y21736118(6x5)
36y217=2(6x5)
36y217=12x+10
36y+12x227=0
Hence, the equation of the tangent line to the given curve
(which is perpendicular to line 5y15x=13) is 36y+12x227=0

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Similar Questions
Q1
Find the equation of the tangent line to the curve y=x22x+7 which is.
(a) parallel to the line 2xy+9=0.
(b) perpendicular to the line 5y15x=13.
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Q2
Find the equation of the tangent line to the curve y = x2 − 2x + 7 which is (i) parallel to the line 2x − y + 9 = 0 (ii) perpendicular to the line 5y − 15x = 13.
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Q3
Find the equation of the tangent line to the curve y=x22x+7 which is perpendicular to the line 5y15x=13.
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Q4

Find the equation of the tangent line to the curve y=x22x+7 which is

(a) parallel to the line 2x-y+9=0.

(a) parallel to the line 5y-15x=13.

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Q5
The equation of the tangent line to the curve y=x22x+7 which is perpendicular to the line 5y15x=13 is 12x+36y227=0.If true enter 1 else 0.
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