Gujarati
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

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