logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Find the equation of the normal to curve \(y^2 = 4x\) which passes through the point \((1, 2).\)

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
The curve is $y^2=4x$
Differentiating with respect to $x$
$2y\large\frac{dy}{dx}$$=4$
$\large\frac{dy}{dx}=\frac{4}{2y}$
$\large\frac{dy}{dx}=\frac{2}{y}$
Step 2:
At $(1,2)$
$\large\frac{dy}{dx}=\frac{2}{2}$$=1$
Slope of normal $=-1$
Equation of normal at (1,2)
$y-2=(-1)(x-1)$
(or)$y-2=-x+1$
$x+y-3=0$
answered Aug 9, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...