JEEometry :5

Taking the partial differential of the given equation w.r.t. $x$ and $y$ respectively we get:- $28x-4y-36=0$ $-4x+22y+48=0$ Solving these two we get $x=1, y=-2$ We know that the given equation is a conic and from the above procedure we know that the center of conic lies on the point $(1,-2)$ . Transforming the equation so that center of conic coincides with origin, we get $14(x+1)^{2} -4(x+1)(y-2) + 11(y-2)^{2} -36(x+1) + 48(y-2)+ 41=0$ On simplifying we get:- $14x^{2} + 11y^{2} -4xy = 25$ Rotating the axis by unknown $\theta$ we replace:- $x \to x\cos\theta + y\sin\theta$ $y \to x\sin\theta - y \cos\theta$ Plugging in above equation and simplifying we get:- $x^{2} (14 \cos^{2}\theta + 11\sin^{2} \theta -4\sin\theta\cos\theta) + y^{2}(14 \sin^{2} \theta + 11 \cos^{2}\theta + 4\sin\theta\cos\theta) + xy(6\sin\theta \cos\theta + 4\cos^{2} \theta-4\sin^{2} \theta) = 25$ Since we want $xy$ term to vanish, we have:- $3\sin2\theta + 4\cos2\theta = 0$ Hence $\tan(2\theta) = \frac{-4}{3}$ So we get $\sin2\theta = \frac{-4}{5} , \cos2\theta = \frac{3}{5}$ Hence our equation becomes :- $15x^{2} + 10y^{2} =25$ $\dfrac{3}{5} x^{2} + \dfrac{2}{5} y^{2} =1$