The equation of the director circle of the hy | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d) Equation of &lsquo;director-circle&rsquo; of hyperbola is ${x^2} + {y^2} = {a^2} - {b^2}$.

Here ${a^2} = 16,\,{b^2} = 4$

$	herefore $ ${x^2

Vedclass · Accepted Answer

${x^2} + {y^2} = 12$

Vedclass · Answer

(d) Equation of &lsquo;director-circle&rsquo; of hyperbola is ${x^2} + {y^2} = {a^2} - {b^2}$.

Here ${a^2} = 16,\,{b^2} = 4$

$	herefore $ ${x^2} + {y^2} = 12$ is the required &lsquo;director circle&rsquo;.

The equation of the director circle of the hyperbola $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{4} = 1$ is given by

Similar Questions

A hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at origin and transverse axis is along $x$ - axis. The length of its transverse axis is

The difference of the focal distance of any point on the hyperbola $9{x^2} - 16{y^2} = 144$, is

The eccentricity of the hyperbola can never be equal to

The number of possible tangents which can be drawn to the curve $4x^2 - 9y^2 = 36$ , which are perpendicular to the straight line $5x + 2y -10 = 0$ is

Curve $xy = {c^2}$ is said to be