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If $\frac{x^{2}}{36}-\frac{y^{2}}{k^{2}}=1$ is a hyperbola, then which of the following statements can be true?
Options:
Solution:
1450 Upvotes
Verified Answer
The correct answer is:
$(10,4)$ lies on the hyperbola.
We have,
$$
\frac{x^{2}}{36}-\frac{y^{8}}{k^{2}}=1
$$
Option (a) $(3,1)$
$$
\begin{aligned}
\frac{9}{36}-\frac{1}{k^{2}} &=1 \\
\Rightarrow \quad & \frac{1}{k^{2}}=\frac{-27}{36} \Rightarrow k^{2}=\frac{-36}{27}
\end{aligned}
$$
which is not possible.
Option (b) $(-3,1)$
$$
\frac{9}{36}-\frac{1}{k^{2}}=1 \Rightarrow k^{2}=\frac{-36}{27}
$$
which is not possible.
Option (c) $(5,2)$
$$
\begin{aligned}
\frac{25}{36}-\frac{4}{k^{2}} &=1 \\
\frac{4}{k^{2}} &=\frac{-11}{36} \\
k^{2} &=\frac{-144}{11}
\end{aligned}
$$
which is not possible.
Option (d) $(10,4)$
$$
\frac{100}{36}-\frac{16}{k^{2}}=1 \Rightarrow \frac{16}{k^{2}}=\frac{64}{36}
$$
$\Rightarrow k^{2}=9$, which is possible.
$$
\frac{x^{2}}{36}-\frac{y^{8}}{k^{2}}=1
$$
Option (a) $(3,1)$
$$
\begin{aligned}
\frac{9}{36}-\frac{1}{k^{2}} &=1 \\
\Rightarrow \quad & \frac{1}{k^{2}}=\frac{-27}{36} \Rightarrow k^{2}=\frac{-36}{27}
\end{aligned}
$$
which is not possible.
Option (b) $(-3,1)$
$$
\frac{9}{36}-\frac{1}{k^{2}}=1 \Rightarrow k^{2}=\frac{-36}{27}
$$
which is not possible.
Option (c) $(5,2)$
$$
\begin{aligned}
\frac{25}{36}-\frac{4}{k^{2}} &=1 \\
\frac{4}{k^{2}} &=\frac{-11}{36} \\
k^{2} &=\frac{-144}{11}
\end{aligned}
$$
which is not possible.
Option (d) $(10,4)$
$$
\frac{100}{36}-\frac{16}{k^{2}}=1 \Rightarrow \frac{16}{k^{2}}=\frac{64}{36}
$$
$\Rightarrow k^{2}=9$, which is possible.
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