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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the equation of tangent lines to the curve \( y=4x^3-3x+5\) which are perpendicular to the line \( 9y+x+3=0\)

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  • Equation of the tangent is $y-y_1=m(x-x_1)$
Step 1:
$y=4x^3-3x+5$
On differentiating with respect to $x$
$\large\frac{dy}{dx}=$$12x-3$-------(1)
It is $\perp$ to the line $9y+x+3=0$
The slope of the line is $-(\large\frac{1}{9})$
$\therefore$ The slope of the tangent is $m_1(\large\frac{-1}{9})=$$-1$
$\Rightarrow m_1=9$
$\therefore \large\frac{dy}{dx}$$=9$-----(2)
Equating equ(1) and equ(2)
$12x-3=9$
$12x=12$
$x=1$
If $x=1$,$y=4(1)^3-3(1)+5,y=6$
$\therefore x=1$ and $y=6$
Hence the point of contact of the tangent to the curve is $(x_1,y_1)=(1,6)$
Step 2:
Equation of the tangent is $y-y_1=m(x-x_1)$
$\Rightarrow y-6=9(x-1)$
$\Rightarrow y-6=9x-9$
$9x-y=3$
This is the required equation.
answered Nov 8, 2013 by sreemathi.v
 
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