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If the equation of the tangent drawn at $(h, k)$ to the hyperbola $\frac{(\mathrm{x}-1)^2}{1}-\frac{(\mathrm{y}-2)^2}{2}=1$ is $\mathrm{x}=2$, then $\mathrm{h}+\mathrm{k}=$
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Verified Answer
The correct answer is:
$4$
Given equation of hyperbola is
$\begin{aligned}
& \frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1 ... (i)\\
& \Rightarrow 2(x-1)-(y-2) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2(x-1)}{y-2} ... (ii)
\end{aligned}$
Given equation of tangent is $x=2$, so, slope $=\frac{1}{0}$
$\Rightarrow \frac{2(x-1)}{y-2}=\frac{1}{0} \Rightarrow y=2=k \text { putting in (i), }$
we get $\frac{(\mathrm{x}-1)^2}{1}-0=1 \Rightarrow \mathrm{x}=2=\mathrm{h}$
Now, $h+k=2+2=4$
$\begin{aligned}
& \frac{(x-1)^2}{1}-\frac{(y-2)^2}{2}=1 ... (i)\\
& \Rightarrow 2(x-1)-(y-2) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{2(x-1)}{y-2} ... (ii)
\end{aligned}$
Given equation of tangent is $x=2$, so, slope $=\frac{1}{0}$
$\Rightarrow \frac{2(x-1)}{y-2}=\frac{1}{0} \Rightarrow y=2=k \text { putting in (i), }$
we get $\frac{(\mathrm{x}-1)^2}{1}-0=1 \Rightarrow \mathrm{x}=2=\mathrm{h}$
Now, $h+k=2+2=4$
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