Question

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

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

Answer 1

Answer: (A)

$16x^2-25y^2=400$
$\frac{x^2}{25}-\frac{y^2}{16}=1$
tangents from $(2\sqrt{2},1)$ is $y=mx+c$
$c=1-2m\sqrt{2}$
tangent to the hyperbola in slope form is
$y=mx\pm \sqrt{a^2{m}^2-b^2}$
$c=1-2m\sqrt{2}$
$1-2m\sqrt{2}=\pm \sqrt{a^2{m}^2-b^2}$
square both side
$1+8m^2-4\sqrt{2}m=25m^2-16$
$17m^2+4\sqrt{2}m-17=0$
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.