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Question Number 112159 by bobhans last updated on 06/Sep/20

$$\left(1\right)\mathrm{Find}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{hyperbola}\mathrm{with}$$$$\mathrm{centre}\mathrm{point}\mathrm{at}(1,-2)\mathrm{and}\mathrm{coordinates}$$$$\mathrm{of}\mathrm{foci}\mathrm{is}(6,-2)\mathrm{and}(-4,-2)$$$$\left(2\right)\mathrm{If}\mathrm{hyperbola}\frac{{\mathrm{x}}^2-2\mathrm{nx}+{\mathrm{n}}^2}{25}-\frac{{\mathrm{y}}^2-2\mathrm{my}+{\mathrm{m}}^2}{16}=1$$$$\mathrm{have}\mathrm{a}\mathrm{asympyotes}\mathrm{passes}\mathrm{through}\mathrm{at}$$$$(0,1),\mathrm{then}5\mathrm{m}-4\mathrm{n}=$$

Answered by bemath last updated on 06/Sep/20

$$\left(2\right)\frac{{(x-n)}^2}{5^2}-\frac{{(y-m)}^2}{4^2}=1\to asymptotesy-m=\pm \frac{4}{5}(x-n)$$$$passesthrough(0,1)\to 1-m=\pm \frac{4}{5}(-n)$$$$\to 5-5m=\mp 4n;5m\mp 4n=5$$

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