Question

If a hyperbola passes through the point $P(10,16)$ and it has vertices at $(\pm 6,0)$, then the equation of the normal at $𝑃$ is:

• A. $3x+4y=94$
• B. $x+2y=42$
• C. $2x+5y=100$
• D. $x+3y=58$

Answer 1

The correct option is C

$2x+5y=100$

Explanation of correct option.

Step1. Finding value of $\text{'}b\text{'}$

Given, vertices of hyperbola is $(\pm 6,0)$ and passes through, $(10,16)$

Equation hyperbola is given by, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$here, $a=6$ and hyperbola passes through $(10,16)$

Therefore,

$\begin{array}{rcl}\frac{100}{36}-\frac{256}{b^2}& =& 1\\ -\frac{256}{b^2}& =& 1-\frac{100}{36}\\ b^2& =& 144\\ b& =& 12\end{array}$

Step2. Finding equation of normal:

Therefore equation of normal is given by

$$\begin{gathered} \frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2 \\ \frac{36x}{10}+\frac{144y}{16}=36+144[x_1=10\&y_1=16] \\ \mathrm{divide}\mathrm{whole}\mathrm{equation}\mathrm{by}9 \\ \frac{4\mathrm{x}}{10}+\frac{16\mathrm{y}}{16}=4+16 \end{gathered}$$

On solving,

$2x+5y=100$

Hence, correct option is $\mathbf{\left(}\mathit{C}\mathbf{\right)}$.