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Let the equation of two diameters of a circle $\mathrm{x^{2}+y^{2}-2 x+2 f y+1=0 \text { be } 2 \text { p } x-y=1}$ and $\mathrm{2 x+\mathrm{p} y=4 \mathrm{p}}$ . Then the slope $\mathrm{m \in(0, \infty)}$ of the tangent to the hyperbola $\mathrm{3 x^{2}-y^{2}=3}$ passing through the centre of the circle is equal to _________.
Option: 1
2

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{2 p+f-1=0} \\$ ......(1)

$\mathrm{2-p f-4 p=0} \\$ .......(2)

$\mathrm{2=p(f+4)} \\$

$\mathrm{p=\frac{2}{(f+4)}} \\$

$\mathrm{2 P=(1-f)} \\$

$\mathrm{\frac{4}{f+4}=1-f} \\$

$\mathrm{f^{2}+3 f=0}$

$\mathrm{f=0 or -3}f=0 or -3$

$\mathrm{Hyperbola\; 3 x^{2}-y^{2}=3, \quad x^{2}-\frac{y^{2}}{3}=1}\\$

$\mathrm{y=m x \pm \sqrt{m^{2}-3}}$

$\mathrm{if \;passes (1,0)}$

$\mathrm{0=m \pm \sqrt{m^{2}-3}}$

$\mathrm{m\;tends \infty}$

$\mathrm{it\; passes (1,3)}$

$\mathrm{3=m \pm \sqrt{m^{2}-3}}$

$\mathrm{(3-m)^{2}=m^{2}-3}$

$\mathrm{m=2}$

hence answer is 2

Posted by

Pankaj Sanodiya

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