Question

find the equation of tangent to curve y=4x^2-3x+5 which are perpendicular to the line 9y+x+3=0

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Find the equation of tangent lines to the curve \( y=4x^3-3x+5\) which are perpendicular to the line \( 9y+x+3=0\)

cbse class12 modelpaper 2012 sec-b q16 medium math

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asked Feb 6, 2013 by thanvigandhi_1
retagged Nov 8, 2013 by sreemathi.v

1 Answer

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Equation of the tangent is y−y1=m(x−x1)y−y1=m(x−x1)
Step 1:
y=4x3−3x+5y=4x3−3x+5
On differentiating with respect to xx
dydx=dydx=12x−312x−3-------(1)
It is ⊥⊥ to the line 9y+x+3=09y+x+3=0
The slope of the line is −(19)−(19)
∴∴ The slope of the tangent is m1(−19)=m1(−19)=−1−1
⇒m1=9⇒m1=9
∴dydx∴dydx=9=9-----(2)
Equating equ(1) and equ(2)
12x−3=912x−3=9
12x=1212x=12
x=1x=1
If x=1x=1,y=4(1)3−3(1)+5,y=6y=4(1)3−3(1)+5,y=6
∴x=1∴x=1 and y=6y=6
Hence the point of contact of the tangent to the curve is (x1,y1)=(1,6)(x1,y1)=(1,6)
Step 2:
Equation of the tangent is y−y1=m(x−x1)y−y1=m(x−x1)
⇒y−6=9(x−1)⇒y−6=9(x−1)
⇒y−6=9x−9⇒y−6=9x−9
9x−y=39x−y=3
This is the required equation.