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Find the equation of the tangent and normal to the ellipse $16x^{2}+25y^{2}=400 $ at $t=\large\frac{1}{\sqrt{3}}$

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  • 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:
Tangent and normal at $t=\large\frac{1}{\sqrt 3}$
$t=\tan \large\frac{\theta}{2}=\frac{1}{\sqrt 3}$
Therefore $\large\frac{\theta}{2}=\frac{\pi}{6}$
$\Rightarrow\theta=\large\frac{\pi}{3}$
The equation of the ellipse is $16x^2+25y^2=400$
The above equation is divided by $400$ we get
$\large\frac{x^2}{25}+\frac{y^2}{16}=$$1$
$a^2=24,b^2=16$
$a=5,b=4$
The parametric equations are $x=5\cos\theta$,$y=4\sin\theta$
Step 2:
The tangent at $\theta$ is $\large\frac{x\cos \theta}{a}+\large\frac{y\sin \theta}{b}$$=1$
At $\theta=\large\frac{\pi}{3}$ we get,
$\large\frac{x\cos \Large\frac{\pi}{3}}{5}+\large\frac{y\sin \Large\frac{\pi}{3}}{4}$$=1$
$\large\frac{x}{10}+\large\frac{y\sqrt 3}{8}$$=1$
$4x+5\sqrt 3y=40$
Step 3:
The normal at $\theta$ is given by $\large\frac{ax}{\cos \theta}-\large\frac{by}{\sin \theta}=$$a^2-b^2$
At $\theta=\large\frac{\pi}{3}$ we get,
$\large\frac{5x}{\cos \Large\frac{\pi}{3}}-\large\frac{4y}{\sin \Large\frac{\pi}{3}}=$$25-16$
$\large\frac{5x}{\Large\frac{1}{2}}-\large\frac{4y}{\Large\frac{\sqrt 3}{2}}$$=25-16$
$10x-\large\frac{8y}{\sqrt 3}$$=9$
On simplifying we get,
$10\sqrt 3x-8y-9\sqrt 3=0$
Step 4:
Equation of tangent : $4x+5\sqrt 3y=40$
Equation of normal : $10\sqrt 3x-8y-9\sqrt 3=0$
answered Jun 20, 2013 by sreemathi.v
 

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