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Let the curve $x^2+2 y^2=2$ intersect the line $x+y=1$ at two points $\mathrm{P}$ and $\mathrm{Q}$ and $\mathrm{O}$ be the origin. If $\theta$ is the acute angle between the lines $\mathrm{OP}$ and $\mathrm{OQ}$, then $\tan \theta=$
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Verified Answer
The correct answer is:
$4$
Given the curve $x^2+2 y^2=2 \& x+y=1$
$\begin{aligned}
& x^2+2 y^2=2 \Rightarrow x^2+2 y^2=2(1)^2 \\
& \Rightarrow x^2+2 y^2=2(x+y)^2 \\
& \Rightarrow x^2+2 y^2=2 x^2+2 y^2+4 x y \Rightarrow x^2+4 x y=0
\end{aligned}$
Now, $\tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{2^2-0}}{1+0}\right|=4 \Rightarrow \tan \theta=4$
$\begin{aligned}
& x^2+2 y^2=2 \Rightarrow x^2+2 y^2=2(1)^2 \\
& \Rightarrow x^2+2 y^2=2(x+y)^2 \\
& \Rightarrow x^2+2 y^2=2 x^2+2 y^2+4 x y \Rightarrow x^2+4 x y=0
\end{aligned}$
Now, $\tan \theta=\left|\frac{2 \sqrt{h^2-a b}}{a+b}\right|=\left|\frac{2 \sqrt{2^2-0}}{1+0}\right|=4 \Rightarrow \tan \theta=4$
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