The centre of the conic represented by the eq | vedclass

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(D) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing with $2x^2 - 72xy + 23y^2 - 4x - 28y - 48 = 0$,we have $a=2,

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$\left( -\frac{11}{25}, -\frac{2}{25} \right)$

Vedclass · Answer

(D) The given equation is of the form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$.Comparing with $2x^2 - 72xy + 23y^2 - 4x - 28y - 48 = 0$,we have $a=2, h=-36, b=23, g=-2, f=-14, c=-48$.The centre $(x, y)$ is given by the partial derivatives $\frac{\partial}{\partial x} = 0$ and $\frac{\partial}{\partial y} = 0$.$4x - 72y - 4 = 0 \implies x - 18y = 1$ $(i)$$-72x + 46y - 28 = 0 \implies -36x + 23y = 14$ (ii)Multiplying $(i)$ by $36$: $36x - 648y = 36$.Adding to (ii): $(23 - 648)y = 14 + 36 \implies -625y = 50 \implies y = -\frac{50}{625} = -\frac{2}{25}$.Substituting $y$ in $(i)$: $x - 18(-\frac{2}{25}) = 1 \implies x + \frac{36}{25} = 1 \implies x = 1 - \frac{36}{25} = -\frac{11}{25}$.Thus,the centre is $\left( -\frac{11}{25}, -\frac{2}{25} \right)$.

The centre of the conic represented by the equation $2x^2 - 72xy + 23y^2 - 4x - 28y - 48 = 0$ is

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