Let x^2+y^2=16 be the equation of the auxilia | vedclass

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

Question

(A) The auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$. $	herefore a^2=16 \Rightarrow a=4$. The hyperbola passe

Vedclass · Accepted Answer

$5 / 4$

Vedclass · Answer

(A) The auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2$. $	herefore a^2=16 \Rightarrow a=4$. The hyperbola passes through $(4 \sqrt{2}, 3)$,so: $\frac{(4 \sqrt{2})^2}{16} - \frac{3^2}{b^2} = 1$ $\Rightarrow \frac{32}{16} - \frac{9}{b^2} = 1$ $\Rightarrow 2 - \frac{9}{b^2} = 1$ $\Rightarrow \frac{9}{b^2} = 1$ $\Rightarrow b^2 = 9$. The eccentricity $e$ is given by $e = \sqrt{1 + \frac{b^2}{a^2}}$. $e = \sqrt{1 + \frac{9}{16}} = \sqrt{\frac{25}{16}} = \frac{5}{4}$.

Let $x^2+y^2=16$ be the equation of the auxiliary circle of a hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and let $(4 \sqrt{2}, 3)$ be a point on the hyperbola. Then,the eccentricity of the hyperbola is

Similar Questions

The equation of the normal to the hyperbola $\frac{x^{2}}{16} - \frac{y^{2}}{9} = 1$ at $(-4, 0)$ is

If the equation of the hyperbola having $(8,3)$ and $(0,3)$ as foci and $\frac{4}{3}$ as eccentricity is $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$,then $p+q=$

If the latus rectum of a hyperbola subtends a right angle at the center,then its eccentricity is:

The distance between the directrices of a rectangular hyperbola is $10$ units. What is the distance between its foci?

The locus of the point of intersection of lines $(x + y)t = a$ and $x - y = at$,where $t$ is the parameter,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module