A hyperbola passes through the points (3, 2) | vedclass

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(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(3, 2)$:$\frac{9}{a^2} - \frac{4}{b^2} = 1$

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$2$

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(A) Let the equation of the hyperbola be $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Since it passes through $(3, 2)$:$\frac{9}{a^2} - \frac{4}{b^2} = 1$ ..... $(i)$Since it passes through $(-17, 12)$:$\frac{289}{a^2} - \frac{144}{b^2} = 1$ ..... $(ii)$Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$. The equations become:$9u - 4v = 1$ ..... $(iii)$$289u - 144v = 1$ ..... $(iv)$Multiply $(iii)$ by $36$:$324u - 144v = 36$ ..... $(v)$Subtract $(iv)$ from $(v)$:$(324 - 289)u = 36 - 1$$35u = 35 \implies u = 1 \implies a^2 = 1 \implies a = 1$.Substitute $u=1$ into $(iii)$:$9(1) - 4v = 1 \implies 4v = 8 \implies v = 2 \implies b^2 = \frac{1}{2}$.The length of the transverse axis is $2a = 2(1) = 2$.

$A$ hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at the origin and its transverse axis along the $x$-axis. The length of its transverse axis is:

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