Question

If $\frac{x^2}{36}-\frac{y^2}{k^2}=1$ is a hyperbola, then which of the following statements can be correct?

• A. $(-3,1)$ lies on the hyperbola
• B. $(3,1)$ lies on the hyperbola
• C. $(10,4)$ lies on the hyperbola
• D. $(5,2)$ lies on the hyperbola

Answer 1

The correct option is C

$(10,4)$ lies on the hyperbola

Explanation for the correct option.

The equation of hyperbola is given to be $\frac{x^2}{36}-\frac{y^2}{k^2}=1$.

$$\begin{gathered} \Rightarrow \frac{y^2}{k^2}=\frac{x^2}{36}-1 \\ \Rightarrow \frac{y^2}{k^2}=\frac{x^2-36}{36} \\ \Rightarrow k^2=\frac{36y^2}{x^2-36} \end{gathered}$$

Now, $k^2>0$

Let us check it for, $\left(10,4\right)$

$\begin{array}{rcl}{k}^2& =& \frac{36\left(16\right)}{100-36}\\ & =& 9\end{array}$

Hence, option C is correct.

Explanation for incorrect options.

Option A

$k^2>0$

Let us check it for, $\left(-3,1\right)$

$\begin{array}{rcl}{k}^2& =& \frac{36\left(1\right)}{9-36}\\ & \Rightarrow & k^2<0\end{array}$

Hence, option A is incorrect.

Option B

$k^2>0$

Let us check it for, $\left(3,1\right)$

$\begin{array}{rcl}{k}^2& =& \frac{36\left(1\right)}{9-36}\\ & \Rightarrow & k^2<0\end{array}$

Hence, option B is incorrect.

Option D

$k^2>0$

Let us check it for, $(5,2)$

$\begin{array}{rcl}{k}^2& =& \frac{36\left(4\right)}{25-36}\\ & \Rightarrow & k^2<0\end{array}$

Hence, option D is incorrect.

Hence, the correct answer is option C.