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Q)

Find the equation of the two tangents that can be drawn, from the point $(1 , 2 )$ to the hyperbola $2x^{2}-3y^{2}=6$

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A)
Toolbox:
  • Equation of any tangent to
  • $\quad(i)$ The parabola $y^2=4ax$ is of the form $y=mx+\large\frac{a}{m}$
  • $\quad(ii)$ The ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2+b^2}$
  • $\quad(iii)$ The hyperbola $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{a^2m^2-b^2}$
Step 1:
$2x^2-3y^2=6$
The above equation is divided by $6$
$\large\frac{x^2}{3}-\large\frac{y^2}{2}$$=1$
Therefore $a^2=3,b^2=2$
Any tangent to $\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1$ is of the form $y=mx\pm\sqrt{3m^2-2}$
Let the required tangent to this ellipse be $y=mx+\sqrt{3m^2-2}$
Step 2:
It passes through $(1,2)$
Therefore $2=m+\sqrt{3m^2-2}$
$(2-m)^2=3m^2-2$
$4+m^2-4m=3m^2-2$
$2m^2+4m-6=0$
$m^2+2m-3=0$
$m=\large\frac{-2\pm \sqrt{4+12}}{2}$
$\Rightarrow\large\frac{ -2\pm 4}{2}$$=-3,1$
Step 3:
Using point slope form,since they pass through $(1,2)$
When $m=1$
$y-2=1(x-1)$
$x-y+1=0$
When $m=-3$
$y-2=-3(x-1)$
$(y-2)=-3x+3$
$3x+y-5=0$
Step 4:
The two tangents are $x-y+1=0,3x+y-5=0$
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