- Home
- Standard 11
- Mathematics
10-2. Parabola, Ellipse, Hyperbola
easy
The equation of the director circle of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$ is given by
A
${x^2} + {y^2} = 16$
B
${x^2} + {y^2} = 4$
C
${x^2} + {y^2} = 20$
D
${x^2} + {y^2} = 12$
Solution
(d) Equation of ‘director-circle’ of hyperbola is ${x^2} + {y^2} = {a^2} – {b^2}$.
Here ${a^2} = 16,\,{b^2} = 4$
$\therefore $ ${x^2} + {y^2} = 12$ is the required ‘director circle’.
Standard 11
Mathematics