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Find the equation of the tangent and normal to the ellipse $2x^{2}+3y^{2}=6 $ at $ (\sqrt{3} , 0 )$

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  • The equation of the tangent at $(x_1,y_1)$ to a conic $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ is given by $Axx_1+\large\frac{B}{2}$$(xy_1+x_1y)+Cyy_1+D\large\frac{(x+x_1)}{2}+$$E\large\frac{(y+y_1)}{2}$$+F=0$
  • The normal is the perpendicular to this line that passes through $(x_1,y_1)$
  • Any line perpendicular to $ax+by+c=0$ is of the form $bx-ay+k=0$.This can be used to find the equation of the normal.
Step 1:
Tangent and normal at $(\sqrt 3,0)$ to the ellipse $2x^2+3y^2=6$
The tangent at $(x_1,y_1)$ is $2xx_1+3yy_1=6$
Here $(x_1,y_1)=(\sqrt 3,0)$
The tangent is $2\sqrt 3 x=6$
$\Rightarrow \sqrt 3x-3=0$
$x-\sqrt 3=0$
Step 2:
The normal $\perp$ to the tangent and is of the form $y=k$.It passes through $(\sqrt 3,0)\Rightarrow k=0$.
Normal is $y=0$
Equation of tangent $x=\sqrt 3$
Equation of normal $y=0$
Note : $(\sqrt 3,0)$ can be seen to be one of the vertices of the ellipse.The tangent at the vertex $(a,0)$ is $x=a$.So the tangent is $x=\sqrt 3$).
answered Jun 20, 2013 by sreemathi.v

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