- Home
- Standard 11
- Mathematics
10-2. Parabola, Ellipse, Hyperbola
hard
यदि अतिपरवलय $16 x ^{2}-9 y ^{2}=144$ की नियता (directrix) $5 x+9=0$ है, तो इसका संगत नाभिकेन्द्र है
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