Question

How many real tangents can be drawn from the point $(4,3)$
to hyperbola $\frac{x^2}{16}-\frac{y^2}{9}=1$ Find the equation of these tangents & angle between them.

Answer 1

Given point $P=(4,3)$

Hyperbola $S=\frac{x^2}{16}-\frac{y^2}{9}-1=0\because S_1=\frac{16}{16}-\frac{9}{9}-1=-1<0\Rightarrow$ Point P = (4,3) lies outside the hyperbola.

$\because$ Two tangents can be

drawn from the the point P (4,3).

Equation of pair of tangents is $S{S}_1=T^2\Rightarrow (\frac{x^2}{16}-\frac{y^2}{9}-1)(-1)$

$={(\frac{4x}{16}-\frac{3y}{9}-1)}^2\Rightarrow -\frac{x^2}{16}+\frac{y^2}{9}+1$

$=\frac{x^2}{9}+1-\frac{xy}{6}-\frac{x}{2}+\frac{2y}{3}0={\tan }^{-1}\left(\frac{4}{3}\right)$