A rectangular hyperbola of latus rectum 2 uni | vedclass

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

Question

(B) For a rectangular hyperbola,the eccentricity $e = \sqrt{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 2$,which implies $b^2 = a$.For a r

Vedclass · Accepted Answer

$x^2 + y^2 = 9$

Vedclass · Answer

(B) For a rectangular hyperbola,the eccentricity $e = \sqrt{2}$.The length of the latus rectum is $\frac{2b^2}{a} = 2$,which implies $b^2 = a$.For a rectangular hyperbola,$a^2 = b^2$,so $a^2 = a$,which gives $a = 1$ (since $a > 0$).Thus,$a = 1$ and $b = 1$.The distance between the foci is $2ae = 2(1)(\sqrt{2}) = 2\sqrt{2}$.Let the other focus be $(h, k)$. The given focus is $S_1 = (1, 0)$ and the point on the hyperbola is $P = (0, 0)$.By the definition of a hyperbola,$|PS_1 - PS_2| = 2a$.$|\sqrt{(0-1)^2 + (0-0)^2} - \sqrt{(0-h)^2 + (0-k)^2}| = 2(1)$.$|1 - \sqrt{h^2 + k^2}| = 2$.This implies $\sqrt{h^2 + k^2} = 3$ or $\sqrt{h^2 + k^2} = -1$ (impossible).Therefore,$h^2 + k^2 = 9$,which represents the curve $x^2 + y^2 = 9$.

$A$ rectangular hyperbola of latus rectum $2$ units passes through $(0, 0)$ and has $(1, 0)$ as its one focus. The other focus lies on the curve -

Similar Questions

The eccentricity of the conic ${x^2} - 4{y^2} = 1$ is

Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$. If the normal at $P$ intersects the $X$-axis at $(16,0)$, then the eccentricity of the hyperbola is

The equation of the hyperbola whose coordinates of the foci are $(\pm 8, 0)$ and the length of the latus rectum is $24$ units, is

Find the coordinates of the foci and the vertices,the eccentricity,and the length of the latus rectum of the hyperbola $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$.

If the equation $\frac{x^2}{7 - k} + \frac{y^2}{5 - k} = 1$ represents a hyperbola,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module