# If a hyperbola passes through the point

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 to it at $P$ is:

1. (1) $3 x+4 y=94$

2. (2) $2 x+5 y=100$

3. (3) $x+2 y=42$

4. (4) $x+3 y=58$

Correct Option: , 2

Solution:

Let the hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

If a hyperbola passes through vertices at $(\pm 6,0)$, then

$\therefore \quad a=6$

As hyperbola passes through the point $P(10,16)$

$\therefore \quad \frac{100}{36}-\frac{256}{b^{2}}=1 \Rightarrow b^{2}=144$

$\therefore \quad$ Required hyperbola is $\frac{x^{2}}{36}-\frac{y^{2}}{144}=1$

$\therefore \quad$ Required hyperbola is $\frac{x^{2}}{36}-\frac{y^{2}}{144}=1$

Equation of normal is $\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}$

$\therefore \quad$ At $\mathrm{P}(10,16)$ normal is

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

$\therefore \quad 2 x+5 y=100 .$

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