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Q)

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. Comment
A)
Toolbox:
• 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}$$=1 if c=a^2m^2+b^2 Here the line is x-y+4=0 or y=x+4. Therefore m=1,c=4 The ellipse is x^2+3y^2=12 The above equation is divided by 12 we get \large\frac{x^2}{12}+\large\frac{y^2}{4}$$=1$
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}$$=-3 \large\frac{b^2}{c}=\large\frac{4}{4}$$=1$
The point of contact is $(-3,1)$