logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the equation of the tangent and normal to the ellipse $ x^{2} +4y^{2}=32 $ at $\theta=\large\frac{\pi}{4}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • The equation of tangent at $\theta$ is given by $\large\frac{x\cos \theta}{a}+\large\frac{y\sin \theta}{b}$$=1$
  • The equation of normal at $\theta$ is given by $\large\frac{ax}{\cos \theta}-\large\frac{by}{\sin \theta}=$$a^2-b^2$
Step 1:
$x^2+4y^2=32$
The above equation is divided by $32$ we get
$\large\frac{x^2}{32}+\large\frac{4y^2}{32}$$=1$
$\Rightarrow \large\frac{x^2}{32}+\large\frac{y^2}{8}$$=1$
Here $2a=32$
$a=4\sqrt 2$
$b^2=8$
$b=2\sqrt 2$
The parametric equations are $x=4\sqrt 2\cos\theta$,$y=2\sqrt 2\sin\theta$
Step 2:
The tangent at $\theta$ is $\large\frac{x\cos \theta}{a}+\large\frac{y\sin \theta}{b}$$=1$
At $\theta=\large\frac{\pi}{4}$ we get,
$\large\frac{x\cos \Large\frac{\pi}{4}}{4\sqrt 2}+\large\frac{y\sin \Large\frac{\pi}{4}}{2\sqrt 2}$$=1$
On simplifying we get,
$\large\frac{x}{8}+\frac{y}{4}$$=1$
$x+2y=8$
Step 3:
The normal at $\theta$ is given by $\large\frac{ax}{\cos \theta}-\large\frac{by}{\sin \theta}=$$a^2-b^2$
At $\theta=\large\frac{\pi}{4}$ we get,
$\large\frac{4\sqrt 2x}{\cos \Large\frac{\pi}{4}}-\large\frac{2\sqrt 2y}{\sin \Large\frac{\pi}{4}}=$$32-8$
On simplifying we get,
$8x-4y=24$
This equation is divided by 4 we get
$2x-y=6$
Step 4:
Equation of tangent : $x+2y-8=0$
Equation of normal : $2x-y-6=0$
answered Jun 20, 2013 by sreemathi.v
edited Jun 20, 2013 by sreemathi.v
 

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
...