Find the equation of the hyperbola satisfying the give conditions: Foci $(0,\,\pm 13),$ the conjugate axis is of length $24.$
Foci $(0,\,\pm 13),$ the conjugate axis is of length $24$.
Here, the foci are on the $y-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$
since the foci are $(0,\,\pm 13)$, $c=13$
since the length of the conjugate axis is $24$, $2 b=24 \Rightarrow b=12$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore a^{2}+12^{2}=13^{2}$
$\Rightarrow a^{2}=169-144=25$
Thus, the equation of the hyperbola is $\frac{y^{2}}{25}-\frac{x^{2}}{144}=1$
If area of quadrilateral formed by tangents drawn at ends of latus rectum of hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is equal to square of distance between centre and one focus of hyperbola, then $e^3$ is ($e$ is eccentricity of hyperbola)
The locus of a point $P(\alpha ,\,\beta )$ moving under the condition that the line $y = \alpha x + \beta $ is a tangent to the hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ is
Let $a$ and $b$ be positive real numbers such that $a > 1$ and $b < a$. Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$. Suppose the tangent to the hyperbola at $P$ passes through the point $(1,0)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$, the normal at $P$ and the $x$-axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are $TRUE$?
$(A)$ $1 < e < \sqrt{2}$
$(B)$ $\sqrt{2} < e < 2$
$(C)$ $\Delta=a^4$
$(D)$ $\Delta=b^4$
If angle between asymptotes of hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{3} = 4$ is $\frac {\pi }{3}$ , then its conjugate hyperbola is
For the hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{3} = 1$ the incorrect statement is :