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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of the tangents drawn from the point $(-1,-2)$ to the hyperbola $2x^2-3y^2=6 $

$\begin{array}{1 1}(A)\;x-y+1=0 \; ; \; 3x+y+7=0 \\(B)\;x+y+1=0 \; ; \; 3x-y-7=0 \\(C)\;x+y-1=0 \; ; \; 3x-y+7=0 \\(D)\;none\;of \;these \end{array}$

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1 Answer

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Hyperbola is $\large\frac{x^2}{3}-\frac{y^2}{2}$$=1$
$a^2=3$
$b^2=2$
Any tangent is $y=mx+\sqrt {a^2m^2-b^2}$
It passes through $(-1,-2)$
$-2=-m+ \sqrt {3m^2-2}$
$\qquad= (m-2)^2=3m^2-2$
=> $m^2+2m-3=0$
=> $m=1,3$
Tangent are $x-y+1=0$
$3x+y+7=0$
Hence A is the correct answer.
answered Apr 21, 2014 by meena.p
 

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