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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  <div class='qna-option

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

 

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

\mathrm{b^2=d_1 d_2 \text {, where } d_1 \text { and } d_2}  are perpendicular distances of any tangent from focii of the hyperbola \mathrm{b^2=4 \times 2=8 \quad \Rightarrow \quad b=2 \sqrt{2}}

\mathrm{2 a e=\sqrt{72}=6 \sqrt{2}}

\mathrm{\text { So } \quad a^2=10}

Centre is (5, 1)

Equation of conjugate axis \mathrm{x-y=4}  and equation of transverse axis is \mathrm{x+y=6}

\mathrm{\therefore \quad \text { Equation of hyperbola is } \frac{(x-y-4)^2}{20}-\frac{(x+y-6)^2}{16}=1}

 

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