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The point of contact of the line $y=x-1$ with $3 x^2-4 y^2=12$ is
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Verified Answer
The correct answer is:
$(4,3)$
The equation of the line and hyperbola are
$\begin{aligned}
& y=x-1 ....(i)\\
& 3 x^2-4 y^2=12 ....(ii)
\end{aligned}$
From (i) and (ii), we get
$\begin{aligned}
& 3 x^2-4(x-1)^2=12 \\
& \Rightarrow 3 x^2-4\left(x^2-2 x+1\right)=12
\end{aligned}$
or $x^2-8 x+16=0 \Rightarrow x=4$
From (i), $y=3$. So point of contact is $(4,3)$.
$\begin{aligned}
& y=x-1 ....(i)\\
& 3 x^2-4 y^2=12 ....(ii)
\end{aligned}$
From (i) and (ii), we get
$\begin{aligned}
& 3 x^2-4(x-1)^2=12 \\
& \Rightarrow 3 x^2-4\left(x^2-2 x+1\right)=12
\end{aligned}$
or $x^2-8 x+16=0 \Rightarrow x=4$
From (i), $y=3$. So point of contact is $(4,3)$.
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