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Let the equation of two diameters of a circle $x ^{2}+ y ^{2}$ $-2 x +2 fy +1=0$ be $2 px - y =1$ and $2 x + py =4 p$. Then the slope $m \in(0, \infty)$ of the tangent to the hyperbola $3 x^{2}-y^{2}=3$ passing through the centre of the circle is equal to $......$
$6$
$2$
$4$
$8$
Solution
$2 p+f-1=0$
$2-p f-4 p=0$
$2=p(f+4)$
$p=\frac{2}{f+4}$
$2 p=1-f$
$\frac{4}{f+4}=1-f$
$f^{2}+3 f=0$
$f=0 \text { or }-3$
Hyperbola $3 x ^{2}- y ^{2}=3, x ^{2}-\frac{ y ^{2}}{3}=1$
$y=m x \pm \sqrt{m^{2}-3}$
It passes $(1,0)$
$o=m \pm \sqrt{m^{2}-3}$
$m$ tends $\infty$
$\text { It passes }(1,3)$
$3=m \pm \sqrt{m^{2}-3}$
$(3-m)^{2}=m^{2}-3$
$m=2$
Similar Questions
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: