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

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