Ask Questions, Get Answers

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

Find the points on the curve \( y=x^3\) at which the slope of the tangent is equal to the y - coordinate of the point.

Can you answer this question?

1 Answer

0 votes
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Let $P(x_1,y_1)$ be the required point.
The given curve is $y=x^3$----(1)
Differentiating w.r.t $x$ we get,
The slope of the tangent at $P(x_1,y_1)$ is
Step 2:
It is given that the slope of the tangent is equal to the $y$ coordinate of the point.
Hence $\large\frac{dy}{dx}$$=y_1$------(3)
Equating equ(2) and equ(3) we get,
Also $(x_1,y_1)$ lies on equ(1)
Hence $y_1=x_1^3$----(5)
Step 3:
Now equating equ(4) and equ(5)
$\Rightarrow 3x_1^2-x_1^3=0$
Therefore $x_1=0$ and $x_1=3$
Step 4:
When $x_1=0,y_1=(0)^3=0$
When $x_1=3,y_1=(3)^3=27$
Hence the required points are $(0,0)$ and $(3,27)$
answered Nov 11, 2013 by sreemathi.v

Related questions

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