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10-2. Parabola, Ellipse, Hyperbola
hard
જો $5x + 9 = 0$ એ અતિવલય $16x^2 -9y^2 = 144$ ની નિયમિકા હોય તો તેને અનુરૂપ નાભી ...... હોય.
A
$(5, 0)$
B
$\left( {\frac{5}{3},0} \right)$
C
$(-5, 0)$
D
$\left( { - \frac{5}{3},0} \right)$
(JEE MAIN-2019)
Solution
$\frac{{{x^2}}}{9} – \frac{{{y^2}}}{{16}} = 1$
$a = 3,b = 4$ and $e = \sqrt {1 + \frac{{16}}{9}} = \frac{5}{3}$
corresponding focus will be $\left( { – ae,0} \right)$ i.e. $\left( { – 5,0} \right)$.
Standard 11
Mathematics