logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
0 votes

Find the equation of normal for the circle, $x^2+y^2+2gx+2fy+c=0$ at $(x_1,y_1)$

$\begin{array}{1 1}(a)\;\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}\\(b)\;\large\frac{(y-y_1)}{(f-y_1)}=\frac{(x+x_1)}{(g-x_1)}\\(c)\;\large\frac{(2y-y_1)}{(f+y_1)}=\frac{(2x-x_1)}{(g+x_1)}\\(d)\;\text{None of these}\end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
We know slope of tangent =$-\large\frac{(g+x_1)}{(f+y_1)}$
Hence slope of normal =$-\large\frac{f+y_1}{g+x_1}$
Hence equation of normal
$(y-y_1)=\large\frac{f+y_1}{g+x_1}$$(x-x_1)$
$\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}$
Hence (a) is the correct answer.
or
Given equation of the circle is
$x^2+y^2+2gx+2fy+c=0$
Differentiating both the sides with respect to $x$ we get
$2x+2y.\large\frac{dy}{dx}$$+2g+2f.\large\frac{dy}{dx}$$=0$
$\Rightarrow\:\large\frac{dy}{dx}=-\large\frac{x+g}{y+f}$
Slope of tangent at any point $(x_1,y_1)$ to any curve is
$\large\frac{dy}{dx}$ at $(x_1,y_1)$
$i.e.,$ Slope of tangent is - $\large\frac{x_1+g}{y_1+f}$
Normal is $\perp$ to tangent
$\therefore$ Slope of normal is $-\large\frac{1}{dy/dx}$
$=\large\frac{y_1+f}{x_1+g}$
$\therefore$ Equation of normal at $(x_1,y_1)$ is
$y-y_1=\large\frac{y_1+f}{x_1+g}$$(x-x_1)$
$\Rightarrow\:\large\frac{(y-y_1)}{(f+y_1)}=\frac{(x-x_1)}{(g+x_1)}$
Hence (a) is the correct answer.
answered Feb 11, 2014 by sreemathi.v
edited Mar 19, 2014 by rvidyagovindarajan_1
 

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
...