Cho hypebol \(\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1\quad (H)\). Khi đó:

(C) có tâm \(I(2;2),R = 2\)

Gọi \(\Delta \) là đường thẳng đi qua \(M(4;6)\).

\(\Delta \) có dạng \(Ax + By - 4A - 6B = 0\)

\(\Delta \) là tiếp tuyến của \((C) \Rightarrow d[I,\Delta ] = R\)

\( \Leftrightarrow \frac{{|2A + 2B - 4A - 6B|}}{{\sqrt {{A^2} + {B^2}} }} = 2 \Leftrightarrow |2A + 4B| = 2\sqrt {{A^2} + {B^2}} \)\(\)

\(\begin{array}{l} \Leftrightarrow 4{A^2} + 16{B^2} + 16AB = 4{A^2} + 4{B^2} \Leftrightarrow 12{A^2} + 16AB = 0 \Leftrightarrow 4B(3B + 4A) = 0\\ \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{B = 0}\\{B = \frac{{4A}}{3}}\end{array}{\rm{.  }}} \right.\end{array}\)

Chọn \(A = 3 \Rightarrow \left[ {\begin{array}{*{20}{l}}{B = 0}\\{B = 4}\end{array}.} \right.\)

Vậy \(A = 3,B = 0 \Rightarrow \left( {{\Delta _1}} \right):x - 4 = 0;A = 3,B = 4 \Rightarrow \left( {{\Delta _2}} \right):3x + 4y + 12 = 0\).

\((E):\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {{a^2} = {b^2} + {c^2};a,b,c > 0} \right) \cdot \left\{ {\begin{array}{*{20}{l}}{\frac{c}{a} = \frac{{\sqrt 5 }}{3}}\\{2(2a + 2b) = 20 \Rightarrow {a^2} = 9,{b^2} = 4.}\\{{c^2} = {a^2} - {b^2}}\end{array}} \right.\)

\((E):\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\).