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Conic Sections — MATHS STD 11 Science — Question

Rajasthan BoardEnglish MediumSTD 11 ScienceMATHSConic Sections2 Marks
Question
Find the equation of hyperbola which has Vertices $( \pm 7,0),e = \frac{4}{3}$
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Answer

Here vertices are (± 7, 0) which lie on x-axis.
So the equation of hyperbola in standard form is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore$ Vertices (± a, 0) is (± 7, 0) ⇒ a = 7
Now $e = \frac{4}{3} \Rightarrow \frac{c}{a} = \frac{4}{3} \Rightarrow \frac{c}{7} = \frac{4}{3} \Rightarrow c = \frac{{28}}{3}$
We know that c2 = a2 + b2
$\therefore {\left( {\frac{{28}}{3}} \right)^2} = {(7)^2} + {b^2} \Rightarrow {b^2} = \frac{{784}}{9} - 49 = \frac{{343}}{9}$
Thus required equation of hyperbola is
$\frac{{{x^2}}}{{{{(7)}^2}}} - \frac{{{y^2}}}{{\frac{{343}}{9}}} = 1 \Rightarrow \frac{{{x^2}}}{{49}} - \frac{{{9y^2}}}{{343}} = 1$

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