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Show that the line $x-y+4=0$ is a tangent to the ellipse $x^{2}+3y^{2}=12$ . Find the co-ordinates of the point of contact.

1 Answer

  • For $y=mx+c$ to be a tangent to
  • $\quad(i) \;y^2=4ax,c=\large\frac{a}{m}$ and the point of contact is $\big(\large\frac{a}{m^2},\large\frac{2a}{m}\big)$
  • $\quad(ii)\;\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1,c^2=a^2m^2+b^2$ and the point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{b^2}{c}\big)$
  • $\quad(iii)\;\large\frac{x^2}{a^2}-\frac{y^2}{b^2}$$=1,c^2=a^2m^2-b^2$ and the point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{-b^2}{c}\big)$
Step 1:
The line $y=mx+c$ is tangent to the ellipse $\large\frac{x^2}{a^2}+\large\frac{y^2}{b^2}$$=1$ if $c=a^2m^2+b^2$
Here the line is $x-y+4=0$ or $y=x+4$.
Therefore $m=1,c=4$
The ellipse is $x^2+3y^2=12$
The above equation is divided by $12$ we get
Therefore $a^2=12,b^2=4$
Step 2:
We have $a^2m^2+b^2=12\times 1+4=16$
We can see $c^2=16$
Therefore $c^2=a^2m^2+b^2$ and the line is a tangent to the ellipse.
Step 3:
The point of contact is $\big(\large\frac{-a^2m}{c},\large\frac{b^2}{c}\big)$
$\large\frac{-a^2m}{c}=\large\frac{12\times 1}{4}$$=-3$
The point of contact is $(-3,1)$
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