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If a hyperbola whose foci are S=(2, 4) and S'= (8, –2) touches x-axis, then equation of hyperbola is

Option: 1

\frac{\left ( x+y-6 \right )^{2}}{10}-\frac{\left ( x-y-4 \right )^{2}}{8}= 1


Option: 2

\frac{\left ( x-y-4 \right )^{2}}{10}-\frac{\left ( x+y-6 \right )^{2}}{8}= 1


Option: 3

\frac{\left ( x-y-4 \right )^{2}}{20}-\frac{\left ( x+y-6 \right )^{2}}{16}= 1


Option: 4

\frac{\left ( x+y-6 \right )^{2}}{20}-\frac{\left ( x-y-4 \right )^{2}}{16}= 1


Answers (1)

best_answer

As we learned

Coordinates of foci -

\left ( \pm \, ae ,o\right )

- wherein

For the Hyperbola

\frac{x^{2}}{a^{2}}- \frac {y^{2}}{b^{2}}= 1

 

 

b^{2}=d_{1}d_{2}, Where d_{1}\: and\: d_{2},  are perpendicular distances of any tangent from focii of the hyperbola

b^{2}= 4\times 2= 8\: \: \: \Rightarrow b= 2\sqrt{2}

2ae= \sqrt{72}= 6\sqrt{2}

So    a^{2}= 10

Centre is (5, 1)

Equation of conjugate axis x-y=4 and equation of transverse axis is x+y=6

Equation of hyperbola is \frac{\left ( x-y-4 \right )^{2}}{20}-\frac{\left ( x+y-6 \right )^{2}}{16}=1

Posted by

Devendra Khairwa

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