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Q)

Find the equation of the two tangents that can be drawn,from the point $(2 , -3 )$ to the parabola $y^{2}=4x $

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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:
Equations of the tangents from $(2,-3)$ to $y^2=4x$.
Compare with $y^2=4ax$,we have $4a=4\Rightarrow a=1$
Any tangent is of the form $y=mx\pm \large\frac{a}{m}$
Here $a=1$
Therefore $y=mx+\large\frac{1}{m}$
Step 2:
The tangent passes through $(2,-3)$
Therefore $-3=2m+\large\frac{1}{m}$
$2m^2+3m+1=0$
$m=\large\frac{-3\pm\sqrt{9-8}}{4}=\large\frac{-3\pm 1}{4}$$=-1,\large\frac{-1}{2}$
Step 3:
Using the point slope form (since the tangents pass through $(2,-3))$
When $m=-1$
$(y+3)=-1(x-2)$
$(y+3)=-x+2$
$x+y+1=0$
When $m=\large\frac{-1}{2}$
$(y+3)=\large\frac{-1}{2}$$(x-2)$
$2y+6=-x+2$
$\Rightarrow x+2y+4=0$
Step 4:
The two tangents are : $x+y+1=0,x+2y+4=0$
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