Question

The tangents from a point $(2 \sqrt{2}, 1)$ to the hyperbola $16 x ^{2}-25 y ^{2}=400$ include an angle equal to.

• A. $\pi /2$
• B. $\pi /4$
• C. $\pi$
• D. $\pi/3$

Answer 1

Answer: (A)

$\pi /2$

Answer 2

$16 x ^{2}-25 y ^{2}=400$
$\frac{ x ^{2}}{25}-\frac{ y ^{2}}{16}=1$
tangents from $(2 \sqrt{2}, 1)$ is $y = mx + c$
$c=1-2 m \sqrt{2}$
tangent to the hyperbola in slope form is
$y=m x \pm \sqrt{a^{2} m^{2}-b^{2}}$
$c=1-2 m \sqrt{2}$
$1-2 m \sqrt{2}=\pm \sqrt{ a ^{2} m ^{2}- b ^{2}}$
square both side
$1+8 m^{2}-4 \sqrt{2} m=25 m^{2}-16$
$17 m^{2}+4 \sqrt{2} m-17=0$
from above equation we will get slope of tangents $m_{1}, m_{2}$
but before solving this we can see that $m _{1} m _{2}=-1$
it means tangents are perpendicular to each other.