The midpoint of the chord 4x - 3y = 5 of the | vedclass

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(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and

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$(2, 1)$

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(B) The equation of the chord of the hyperbola $S: 2x^2 - 3y^2 - 12 = 0$ with midpoint $M(h, k)$ is given by $T = S_1$,where $T = 2xh - 3yk - 12$ and $S_1 = 2h^2 - 3k^2 - 12$. Thus,the equation of the chord is $2xh - 3yk = 2h^2 - 3k^2$. Comparing this with the given chord equation $4x - 3y = 5$,we get: $\frac{2h}{4} = \frac{-3k}{-3} = \frac{2h^2 - 3k^2}{5}$. From $\frac{2h}{4} = k$,we have $k = \frac{h}{2}$. Substituting $k = \frac{h}{2}$ into $\frac{2h}{4} = \frac{2h^2 - 3(h/2)^2}{5}$: $\frac{h}{2} = \frac{2h^2 - \frac{3h^2}{4}}{5} = \frac{5h^2}{20} = \frac{h^2}{4}$. So,$\frac{h}{2} = \frac{h^2}{4}$ $\Rightarrow 2h = h^2$ $\Rightarrow h(h - 2) = 0$. Since the chord exists,$h 
eq 0$ (as $h=0$ gives $k=0$,which does not satisfy the chord equation $4(0)-3(0)=5$). Thus,$h = 2$. Then $k = \frac{2}{2} = 1$. The midpoint is $(2, 1)$.

The midpoint of the chord $4x - 3y = 5$ of the hyperbola $2x^2 - 3y^2 = 12$ is

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