Let $H : \frac{ x ^2}{ a ^2}-\frac{ y ^2}{ b ^2}=1$, where $a > b >0$, be $a$ hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^{\circ}$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4 \sqrt{3}$..
List $I$ | List $II$ |
$P$ The length of the conjugate axis of $H$ is | $1$ $8$ |
$Q$ The eccentricity of $H$ is | $2$ ${\frac{4}{\sqrt{3}}}$ |
$R$ The distance between the foci of $H$ is | $3$ ${\frac{2}{\sqrt{3}}}$ |
$S$ The length of the latus rectum of $H$ is | $4$ $4$ |
The correct option is:
$P \rightarrow 4 ; Q \rightarrow 2 ; R \rightarrow 1 ; S \rightarrow 3$
$P \rightarrow 4 ; Q \rightarrow 3 ; R \rightarrow 1 ; S \rightarrow 2$
$P \rightarrow 4 ; Q \rightarrow 1 ; R \rightarrow 3 ; S \rightarrow 2$
$P \rightarrow 3 ; Q \rightarrow 4 ; R \rightarrow 2 ; S \rightarrow 1$
A hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at origin and transverse axis is along $x$ - axis. The length of its transverse axis 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
The locus of the point of intersection of the lines $(\sqrt{3}) kx + ky -4 \sqrt{3}=0$ and $\sqrt{3} x-y-4(\sqrt{3}) k=0$ is a conic, whose eccentricity is .............
The equation of the tangent parallel to $y - x + 5 = 0$ drawn to $\frac{{{x^2}}}{3} - \frac{{{y^2}}}{2} = 1$ is