Find the equations and length of transverse and conjugate axes of the hyperbola $\;$ $16x^2 - 9y^2 + 96x + 36y - 36 = 0$
Equation of given hyperbola is: $\;$ $16x^2 - 9y^2 + 96 x + 36 y - 36 = 0$
i.e. $\;$ $16 \left(x^2 + 6x\right) - 9 \left(y^2 - 4y\right) = 36$
i.e. $\;$ $16 \left[\left(x^2 + 6x + 9\right) - 9\right] - 9 \left[\left(y^2 - 4y + 4\right) - 4\right] = 36$
i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144 - 36 + 36$
i.e. $\;$ $16 \left(x + 3\right)^2 - 9 \left(y - 2\right)^2 = 144$
i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{144 / 16} - \dfrac{\left(y - 2\right)^2}{144 / 9} = 1$
i.e. $\;$ $\dfrac{\left(x + 3\right)^2}{9} - \dfrac{\left(y - 2\right)^2}{16} = 1$ $\;\;\; \cdots \; (1)$
Let $X = x + 3$ $\;\;\; \cdots \; (2a)$ $\;$ and $\;$ $Y = y - 2$ $\;\;\; \cdots \; (2b)$
Then in view of equations $(2a)$ and $(2b)$, equation $(1)$ becomes
$\dfrac{X^2}{9} - \dfrac{Y^2}{16} = 1$ $\;\;\; \cdots \; (3)$
The transverse axis of the hyperbola given by equation $(3)$ is along the X axis.
Comparing equation $(3)$ with the standard equation of hyperbola $\;$ $\dfrac{X^2}{a^2} - \dfrac{Y^2}{b^2} = 1$ $\;$ gives
$a^2 = 9 \implies a = 3$; $\;$ $b^2 = 16 \implies b = 4$
| Referred to X, Y | Referred to x, y $X = x + 3$; $\;$ $Y = y - 2$ |
|
|---|---|---|
| Equation of transverse axis | X axis $\;$ $\left(Y = 0\right)$ | $Y = 0 \implies y - 2 = 0$ |
| Equation of conjugate axis | Y axis $\;$ $\left(X = 0\right)$ | $X = 0 \implies x + 3 = 0$ |
| Length of transverse axis | $2a = 2 \times 3 = 6$ | $6$ |
| Length of conjugate axis | $2b = 2 \times 4 = 8$ | $8$ |