Question

Consider the hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$

Which of the following is/are incorrect statement(s) for above hyperbola?

• A. The acute angle between its asymptotes is ${60}^{\circ }$
• B. Length of latus rectum is $\frac{9}{4}$
• C. Eccentricity is $\frac{5}{4}$
• D. Product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than the length of its latus rectum.

Answer 1

Answer: (A), (B), (D)

Length of latus rectum is $\frac{9}{4}$
The acute angle between its asymptotes is ${60}^{\circ }$
Product of the perpendicular distances from any point on the hyperbola on its asymptotes is less than the length of its latus rectum.

Given hyperbola

$H:\frac{x^2}{16}-\frac{y^2}{9}=1$

Eccentricity: $e^2=\frac{b^2}{a^2}+1=\frac{9}{16}+1$

So, $e=\frac{5}{4}$

Length of latus rectum $=\frac{2b^2}{a}=\frac{18}{4}=\frac{9}{2}$

The equations of asymptotes are given by $y=\pm \frac{b}{a}x$
The product of perpendicular distances of any point from asymptotes $=\frac{a^2{b}^2}{a^2+b^2}=\frac{144}{25}$

It is greater than length of the latus rectum.

The slopes of asymptotes are $\pm \frac{b}{a}$ as mentioned earlier, and so if the acute angle between two is $\theta$, then $\tan \theta =\left|\frac{2ab}{a^2-b^2}\right|=\frac{24}{7}$

Hence, the acute angle between them is not ${60}^o$

Hence, option A, B, and D are incorrect.