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

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(B) The equation of the chord of a hyperbola $S = 0$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T$ is the tangent at $(x_1, y_1)$ and $S_

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$(2, 1)$

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(B) The equation of the chord of a hyperbola $S = 0$ with midpoint $(x_1, y_1)$ is given by $T = S_1$,where $T$ is the tangent at $(x_1, y_1)$ and $S_1$ is the value of the hyperbola equation at $(x_1, y_1)$.For the hyperbola $2x^2 - 3y^2 - 12 = 0$,the equation of the chord with midpoint $(x_1, y_1)$ is $2xx_1 - 3yy_1 - 12 = 2x_1^2 - 3y_1^2 - 12$,which simplifies to $2xx_1 - 3yy_1 = 2x_1^2 - 3y_1^2$.Comparing this with the given chord equation $4x - 3y = 5$,we have:$\frac{2x_1}{4} = \frac{-3y_1}{-3} = \frac{2x_1^2 - 3y_1^2}{5}$.From $\frac{2x_1}{4} = y_1$,we get $x_1 = 2y_1$.Substituting $x_1 = 2y_1$ into $\frac{2x_1}{4} = \frac{2x_1^2 - 3y_1^2}{5}$:$\frac{y_1}{1} = \frac{2(2y_1)^2 - 3y_1^2}{5}$ $\Rightarrow 5y_1 = 8y_1^2 - 3y_1^2$ $\Rightarrow 5y_1 = 5y_1^2$.This gives $y_1^2 - y_1 = 0$,so $y_1 = 0$ or $y_1 = 1$.If $y_1 = 0$,then $x_1 = 0$. Checking the point $(0, 0)$ in the chord equation $4(0) - 3(0) = 0 
eq 5$. So,$y_1 = 0$ is rejected.If $y_1 = 1$,then $x_1 = 2(1) = 2$. The point $(2, 1)$ satisfies the chord equation $4(2) - 3(1) = 8 - 3 = 5$.Thus,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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