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Fruendlich Isotherm

Show that the...

Question

Show that the locus of the poles of chords which subtend a constant angle $α$ at the vertex is the curve
$(x + 4 a)^{2} = 4 c o t^{2} a (y^{2} - 4 a x)$

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Solution

For the parabola $y^{2} = 4 a x$ , chord joining points $(a t_{1}^{2}, 2 a t_{1})$ and $(a t_{2}^{2}, 2 a t_{2})$ has the equation $y - 2 a t_{2} = \frac{2 a t_{2} - 2 a t_{1}}{a t_{2}^{2} - a t_{1}^{2}} \times (x - a t_{2}^{2})$
i.e. $(y - 2 a t_{2}) (t_{1} + t_{2}) = 2 (x - a t_{2}^{2})$
Since the chord subtends angle $a$ at the vertex, $\frac{\frac{2}{t_{2}} - \frac{2}{t_{1}}}{1 + \frac{2}{t_{2}} \times \frac{2}{t_{1}}} = tan a$
$∴ 2 (t_{1} - t_{2}) = tan a \times (t_{1} t_{2} + 4) . . . (1)$
Comparing the equation $y k = 2 a x + 2 a h$ with the equation of the chord, we have
$\frac{t_{1} + t_{2}}{k} = \frac{2}{2 a} = \frac{2 a t_{1} t_{2}}{2 a h}$
$\Rightarrow h = a t_{1} t_{2}, k = a (t_{1} + t_{2})$
$\Rightarrow$ Equation (1) becomes $\frac{2}{a} \times \sqrt{k^{2} - 4 a h} = tan a \times (\frac{h}{a} + 4)$
i.e. $(x + 4 a)^{2} = 4 {cot}^{2} a (y^{2} - 4 a x)$ becomes the required locus

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