Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Math  >>  Model Papers
0 votes

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

Can you answer this question?

1 Answer

0 votes
  • Equation of the tangent is $y-y_1=m(x-x_1)$
Step 1:
On differentiating with respect to $x$
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)
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 $y-y_1=m(x-x_1)$
$\Rightarrow y-6=9(x-1)$
$\Rightarrow y-6=9x-9$
This is the required equation.
answered Nov 8, 2013 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App