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  4. The line y=x-1 touches the curve 3 x2-4 y2=12. The point of contact is

Q. The line $y=x-1$ touches the curve $3 x^{2}-4 y^{2}=12$. The point of contact is

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Solution:

$3 x^{2}-4(x-1)^{2}=12 $
$\Rightarrow 3 x^{2}-4 x^{2}+8 x-4=12$
$ \Rightarrow-x^{2}+8 x-16=0$
$\Rightarrow x^{2}-8 x+16=0$
$(x-4)^{2}=0 \Rightarrow x=4, y=3$
$\therefore$ Point of contact is $(4,3)$

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