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10-2. Parabola, Ellipse, Hyperbola
easy
The length of transverse axis of the parabola $3{x^2} - 4{y^2} = 32$ is
A
$\frac{{8\sqrt 2 }}{{\sqrt 3 }}$
B
$\frac{{16\sqrt 2 }}{{\sqrt 3 }}$
C
$\frac{3}{{32}}$
D
$\frac{{64}}{3}$
Solution
(a) The given equation may be written as $\frac{{{x^2}}}{{32/2}} – \frac{{{y^2}}}{8} = 1$ or $\frac{{{x^2}}}{{{{\left( {4\sqrt 2 /\sqrt 3 } \right)}^2}}} – \frac{{{y^2}}}{{{{(2\sqrt 2 )}^2}}} = 1$.
Comparing the given equation with $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$,
we get ${a^2} = {\left( {\frac{{4\sqrt 2 }}{{\sqrt 3 }}} \right)^2}$ or $a = \frac{{4\sqrt 2 }}{{\sqrt 3 }}.$
Therefore length of transverse axis of a hyperbola
$ = 2a = 2 \times \frac{{4\sqrt 2 }}{{\sqrt 3 }} = \frac{{8\sqrt 2 }}{{\sqrt 3 }}.$
Standard 11
Mathematics
