Question

The slope of the tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{16}=1$ which passes through the point $(1,4)$ is

• A. $-1$
• B. $-\frac{4}{3}$
• C. $2$
• D. $1$

Answer 1

Answer: (B), (D)

$1$
$-\frac{4}{3}$

The equation of a tangent is $y=mx\pm \sqrt{a^2{m}^2-b^2}$
Hence, $y=mx\pm \sqrt{25m^2-16}$ is the equation of a tangent to the hyperbola.
This passes through $(1,4)$, so $4=m\pm \sqrt{25m^2-16}$
$\Rightarrow 16-8m+m^2=25m^2-16$
$\Rightarrow 24m^2+8m-32=0$
$\Rightarrow m=1,\frac{-4}{3}$