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Q)

Find the equation of the two tangents that can be drawn, from the point $(1 , 3 )$ to the ellipse $4x^{2}+9y^{2}=36$

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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:
$4x^2+9y^2=36$
$\large\frac{x^2}{9}+\frac{y^2}{4}$$=1$
Here $a^2=9,b^2=4$
Let tangent be of the form $y=mx+\sqrt{a^2m^2+b^2}$
(i.e) $y=mx+\sqrt{9m^2+4}$
Step 2:
The tangent passes through $(1,3)$
Therefore $3=m+\sqrt{9m^2+4}$
$(3-m)^2=9m^2+4$
Therefore $9+m^2-6m=9m^2+4$
$\Rightarrow 8m^2+6m-5=0$
$m=\large\frac{-6\pm\sqrt{36+160}}{16}$
$\Rightarrow \large\frac{-6\pm14}{16}=\large\frac{-20}{16},\large\frac{8}{16}$
(i.e) $\Rightarrow \large\frac{-5}{4},\frac{1}{2}$
Step 3:
The two tangents pass through $(1,3)$.
When $m=\large\frac{1}{2}$
$(y-3)=\large\frac{1}{2}$$(x-1)$
$2y-6=x-1$
$x-2y+5=0$
When $m=\large\frac{-5}{4}$
$(y-3)=\large\frac{-5}{4}$$(x-1)$
$4y-12=-5x+5$
$5x+4y-17=0$
Step 4:
The two tangents are $x-2y+5=0,5x+4y-17=0$
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