The distance between the foci of a hyperbola | vedclass

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

Question

(A) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which simplifies to $a = \fr

Vedclass · Accepted Answer

$x^{2}-y^{2}=32$

Vedclass · Answer

(A) The distance between the foci of a hyperbola is given by $2ae = 16$. Given $e = \sqrt{2}$,we have $2a(\sqrt{2}) = 16$,which simplifies to $a = \frac{8}{\sqrt{2}} = 4\sqrt{2}$. Thus,$a^{2} = (4\sqrt{2})^{2} = 32$. For a hyperbola,$b^{2} = a^{2}(e^{2} - 1)$. Substituting the values,$b^{2} = 32((\sqrt{2})^{2} - 1) = 32(2 - 1) = 32$. The standard equation of the hyperbola is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$. Substituting $a^{2} = 32$ and $b^{2} = 32$,we get $\frac{x^{2}}{32} - \frac{y^{2}}{32} = 1$,which simplifies to $x^{2} - y^{2} = 32$.

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

Similar Questions

$A$ metallic square loop $ABCD$ is moving in its own plane with velocity $v$ in a uniform magnetic field perpendicular to its plane as shown in the figure. Electric field is induced:

If $R = \{ (x, y) | x, y \in Z, x^2 + y^2 \le 4 \}$ is a relation in $Z$,then the domain of $R$ is

$AB$ crystallizes in a body-centred cubic lattice with edge length '$a$' equal to $387 \ pm$. The distance between two oppositely charged ions in the lattice is ............. $pm$.

Which of the following is an example of a biofertilizer?

Pure $Si$ at $500\, K$ has equal number of electron $(n_e)$ and hole $(n_h)$ concentrations of $1.5 \times 10^{16} \,m^{-3}$. Doping by indium increases $n_h$ to $4.5 \times 10^{22} \,m^{-3}$. The doped semiconductor is of

Vedclass Test Series

Exam Paper Generator

Online Exam Module