Conics again

If the centre of any conic is at the origin then the coefficient of x and y will be zero in its general equation .

Any point on ellipse is $(r\cos \theta,r\sin \theta)$

After satisfying these coordinates in the given equation and on simplifying we get,

$3r^2 + 2r^2\sin 2\theta =1$

$r= \frac{1}{\sqrt{3 +\sin 2\theta}}$ a = length of semi major axis b = length of semi minor axis

$a=r_{max}=\frac{1}{\sqrt {3-2}}$ [ when denominator is minimum, r wil be maximum and minimum of $\sin 2\theta =-1$

$b=r_{min}=\frac{1}{\sqrt{3+2}}$ { maximum of $\sin 2\theta =1$ ]

Area of ellipse = $\pi ab$

= $\frac{\pi}{\sqrt {5}}$

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$Area ~of ~a~ conics~ having~its~ center~ at~ origin~ is~ of~ the~ form ~~ax^2+2hxy+by^2=1.\\ h^2-ab=4-9<0,~we~ confirm~that~it~is~an~ellipse.~And~so,\\$ $its~area=\dfrac {2\pi}{\sqrt{4ab-(2h)^2}}=\dfrac {\frac {2*22} 7 }{\sqrt{4ab-(2h)^2}}=\dfrac {44}{7*\sqrt{36-16}}=\dfrac{44}{7*2*\sqrt5}=\dfrac{22}{7*\sqrt5}.$