Find the locus of the intersection of tangents which meet at a given angle $α$ .

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Solution

Let the point of intersection be $P (h, k)$

Equation to tangent to $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is

$y = m x \pm \sqrt{a^{2} m^{2} + b^{2}}$

It passes through $(h, k)$

$k = m h \pm \sqrt{a^{2} m^{2} + b^{2}} k - m h = \pm \sqrt{a^{2} m^{2} + b^{2}} k^{2} + m^{2} h^{2} - 2 k m h = a^{2} m^{2} + b^{2} (h^{2} - a^{2}) m^{2} - 2 m h k + k^{2} - b^{2} = 0$

This quadratic in $m$

$m_{1} + m_{2} = \frac{2 h k}{h^{2} - a^{2}} . . . . . . (i) m_{1} m_{2} = \frac{k^{2} - b^{2}}{h^{2} - a^{2}} . . . . . . . (i i)$

$tan α = \frac{m_{1} {- m}_{2}}{1 + m_{1} m_{2}} {tan}^{2} α = \frac{{{(m}_{1} {- m}_{2})}^{2}}{{(1 + m_{1} m_{2})}^{2}} = \frac{{{(m}_{1} {+ m}_{2})}^{2} - 4 m_{1} m_{2}}{{(1 + m_{1} m_{2})}^{2}}$

using $(i)$ and $(i i)$

${tan}^{2} α = \frac{{(\frac{2 h k}{h^{2} - a^{2}})}^{2} - 4 (\frac{k^{2} - b^{2}}{h^{2} - a^{2}})}{{(1 + \frac{k^{2} - b^{2}}{h^{2} - a^{2}})}^{2}} {tan}^{2} α = \frac{\frac{4 h^{2} k^{2} - 4 (k^{2} - b^{2}) (h^{2} - a^{2})}{{(h^{2} - a^{2})}^{2}}}{\frac{{(h^{2} - a^{2} + k^{2} - b^{2})}^{2}}{{(h^{2} - a^{2})}^{2}}}$
${tan}^{2} α = \frac{4 h^{2} k^{2} - 4 (h^{2} k^{2} - h^{2} b^{2} - a^{2} k^{2} + a^{2} b^{2})}{{(h^{2} + k^{2} - a^{2} - b^{2})}^{2}}$
${(h^{2} + k^{2} - a^{2} - b^{2})}^{2} {tan}^{2} α = 4 (h^{2} b^{2} + a^{2} k^{2} - a^{2} b^{2})$
${(h^{2} + k^{2} - a^{2} - b^{2})}^{2} = 4 {cot}^{2} α (h^{2} b^{2} + a^{2} k^{2} - a^{2} b^{2})$

Replacing $h$ by $x$ and $k$ by $y$

${(x^{2} + y^{2} - a^{2} - b^{2})}^{2} = 4 {cot}^{2} α (x^{2} b^{2} + a^{2} y^{2} - a^{2} b^{2})$

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