Ellipse
9.0 Pair of tangents
9.0 Pair of tangents
The combined equation of pair of tangents drawn from a point $P({x_1},{y_1})$ to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ can be find out using
$${T^2} = S{S_1}$$ i.e., $${(\frac{{x{x_1}}}{{{a^2}}} + \frac{{y{y_1}}}{{{b^2}}} - 1)^2} = (\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} - 1)(\frac{{{x_1}^2}}{{{a^2}}} + \frac{{{y_1}^2}}{{{b^2}}} - 1)$$
Question 12. Find the locus of points of intersection of tangents to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ which make an angle $\theta $.
Solution: Let us assume the point of intersection of tangents be $P(\alpha ,\beta )$. Equation of tangent to ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ in terms of slope $(m)$ is $y = mx \pm \sqrt {{a^2}{m^2} + {b^2}} $. Since it pases through point $P$, then $$\beta = m\alpha + \sqrt {{a^2}{m^2} + {b^2}} $$
On squaring both sides, we get $$\begin{equation} \begin{aligned} {(\beta - m\alpha )^2} = {a^2}{m^2} + {b^2} \\ {m^2}({a^2} - {\alpha ^2}) + 2\alpha \beta m + ({b^2} - {\beta ^2}) = 0 \\\end{aligned} \end{equation} $$
which is the quadratic equation in $m$. Let the roots be ${m_1}$ and ${m_2}$, then $$\begin{equation} \begin{aligned} {m_1} + {m_2} = - \frac{{2\alpha \beta }}{{({a^2} - {\alpha ^2})}}...(1), \\ {m_1}{m_2} = \frac{{{b^2} - {\beta ^2}}}{{{a^2} - {\alpha ^2}}}...(2) \\ \\\end{aligned} \end{equation} $$
From equations $(1)$ and $(2)$,
$$\begin{equation} \begin{aligned} {m_1} - {m_2} = \sqrt {{{({m_1} + {m_2})}^2} - 4{m_1}{m_2}} \\ = \sqrt {\frac{{4{\alpha ^2}{\beta ^2}}}{{{{({a^2} - {\alpha ^2})}^2}}} - \frac{{4({b^2} - {\beta ^2})}}{{{a^2} - {\alpha ^2}}}} \\ = \sqrt {\frac{{4{\alpha ^2}{\beta ^2} - 4({b^2} - {\beta ^2})({a^2} - {\alpha ^2})}}{{{{({a^2} - {\alpha ^2})}^2}}}} \\ = \frac{2}{{({a^2} - {\alpha ^2})}}\sqrt {({a^2}{\beta ^2} + {b^2}{\alpha ^2} - {a^2}{b^2})} \\\end{aligned} \end{equation} $$
$\because $ $\theta $ be the angle between these two tangents, then
$$\begin{equation} \begin{aligned} \tan \theta = \left| {\frac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right| \\ = \left| {\frac{{2\sqrt {{a^2}{\beta ^2} + {b^2}{\alpha ^2} - {a^2}{b^2}} }}{{{a^2} + {b^2} - {\alpha ^2} - {\beta ^2}}}} \right| \\\end{aligned} \end{equation} $$
On squaring both sides, we get
$$\begin{equation} \begin{aligned} {({a^2} + {b^2} - {\alpha ^2} - {\beta ^2})^2}{\tan ^2}\theta = 4({a^2}{\beta ^2} + {b^2}{\alpha ^2} - {a^2}{b^2}) \\ {({\alpha ^2} + {\beta ^2} - {a^2} - {b^2})^2}{\tan ^2}\theta = 4({b^2}{\alpha ^2} + {a^2}{\beta ^2} - {a^2}{b^2}) \\\end{aligned} \end{equation} $$
Therefore, locus of point $P(\alpha ,\beta )$ is
$${({x^2} + {y^2} - {a^2} - {b^2})^2}{\tan ^2}\theta = 4({b^2}{x^2} + {a^2}{y^2} - {a^2}{b^2})$$
