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 $2x+5y=100$

Vertex of hyperbola is $(\pm a,0)\equiv (\pm 6,0)\Rightarrow a=6$
Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\Rightarrow \frac{x^2}{36}-\frac{y^2}{b^2}=1$

As $P(10,16)$ lies on the parabola.

$\frac{100}{36}-\frac{256}{b^2}=1$

$\Rightarrow \frac{64}{36}=\frac{256}{b^2}\Rightarrow b^2=144$
Equation of hyperbola becomes $\frac{x^2}{36}-\frac{y^2}{144}=1$

Equation of normal is $\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

$\Rightarrow \frac{36x}{10}+\frac{144y}{16}=180$

$\Rightarrow \frac{x}{50}+\frac{y}{20}=1\Rightarrow 2x+5y=100$