The vertices of a hyperbola are at (0, 0) and | vedclass

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

Question

(B) The vertices are $(0, 0)$ and $(10, 0)$. The center of the hyperbola is the midpoint of the vertices: $(\frac{0+10}{2}, 0) = (5, 0)$.The distance

Vedclass · Accepted Answer

$\frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1$

Vedclass · Answer

(B) The vertices are $(0, 0)$ and $(10, 0)$. The center of the hyperbola is the midpoint of the vertices: $(\frac{0+10}{2}, 0) = (5, 0)$.The distance between the vertices is $2a = 10$,so $a = 5$.The distance from the center $(5, 0)$ to the focus $(18, 0)$ is $ae = 18 - 5 = 13$.Since $a = 5$,we have $5e = 13$,so $e = \frac{13}{5}$.Using the relation $b^2 = a^2(e^2 - 1)$,we get $b^2 = 25(\frac{169}{25} - 1) = 169 - 25 = 144$,so $b = 12$.The equation of the hyperbola with center $(h, k) = (5, 0)$ is $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$.Substituting the values,we get $\frac{(x - 5)^2}{25} - \frac{y^2}{144} = 1$.

The vertices of a hyperbola are at $(0, 0)$ and $(10, 0)$ and one of its foci is at $(18, 0)$. The equation of the hyperbola is:

Similar Questions

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

The equation of the conic with focus at $(1, -1)$,directrix along $x - y + 1 = 0$ and with eccentricity $e = \sqrt{2}$ is:

If $(1,2)$ is the focus,$x+2y=0$ is the directrix,and $\sqrt{2}$ is the eccentricity of a hyperbola,then the equation of the hyperbola is

If $PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ such that $\triangle OPQ$ is an equilateral triangle,where $O$ is the center of the hyperbola,then the eccentricity $e$ of the hyperbola satisfies:

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{y^{2}}{9}-\frac{x^{2}}{27}=1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module