5.3 Slope form

  Hyperbola

The equations of tangents of slope $m$ to hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$ are $$y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $$

Proof: Let $y=mx+c$ be a tangent to the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1...(1)$$

Substituting the value of $y=mx+c$ in $(1)$, we get

$$\begin{equation} \begin{aligned} \frac{{{x^2}}}{{{a^2}}} - \frac{{{{(mx + c)}^2}}}{{{b^2}}} = 1 \\ ({a^2}{m^2} - {b^2}){x^2} + 2mc{a^2}x + {a^2}({c^2} + {b^2}) = 0 \\\end{aligned} \end{equation} $$ must have equal roots.

$$\begin{equation} \begin{aligned} 4{a^4}{m^2}{c^2} - 4({a^2}{m^2} - {b^2}){a^2}({c^2} + {b^2}) = 0{\text{ }}[\because {B^2} - 4AC = 0] \\ {c^2} = {a^2}{m^2} - {b^2} \\ c = \pm \sqrt {{a^2}{m^2} - {b^2}} \\\end{aligned} \end{equation} $$

Substituting the value of $c$ in $y=mx+c$, we get $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $

Question 5. If ${m_1}$ and ${m_2}$ are the slope of tangent to the hyperbola $\frac{{{x^2}}}{{25}} - \frac{{{y^2}}}{{16}} = 1$ which passes through $(6,2)$. Find the value of ${m_1} + {m_2}$ and ${m_1} \times {m_2}$.

Solution: The equation of tangent to the hyperbola in slope form can be written as $y = mx \pm \sqrt {{a^2}{m^2} - {b^2}} $. From the equation of hyperbola, $a=5$ and $b=4$. Therefore,

$$2 = 6m \pm \sqrt {25{m^2} - 16} $$$$2 - 6m = \pm \sqrt {25{m^2} - 16} $$

Squaring both sides, we get $$4 + 36{m^2} - 24m = 25{m^2} - 16$$$$11{m^2} - 24m + 20 = 0$$

On comparing the above equation with the general quadratic equation, $$a'{x^2} + b'x + c' = 0$$

We can write the sum of roots and product of roots of quadratic equation as $${m_1} + {m_2} = - \frac{{b'}}{{a'}} = \frac{{24}}{{12}}$$ and $${m_1} \times {m_2} = \frac{{c'}}{{a'}} = \frac{{20}}{{11}}$$