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Parametric Form of Tangent: Hyperbola

If the point ...

Question

If the point $(a t_{1}^{2}, 2 a t_{1})$ on the parabola $y^{2} = 4 a x$ be called the point $t_{1}$ , prove that the axis of the second parabola through the four points $t_{1}, t_{2}, t_{3}$ and $t_{4}$ makes with the axis of the first an angle
${cot}^{- 1} (\frac{t_{1} + t_{2} + t_{3} + t_{4}}{4}) .$
Prove also that if two parabolas meet in four points the distances of the centroid of the four points from the axes are proportional to the latera recta.

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Solution

Let the axis of the second parabola be parallel to $y - m x = 0$ , so that its equation may be given by
${(y - m x)}^{2} + 2 g x + 2 f y + c = 0$
Let it meet the first parabola in the point $(a t^{2}, 2 a t)$ ; then on substituting and simplifying we get
$t^{4} m^{2} a^{2} - 4 m a^{2} t^{3} +$ terms of lower degree in $t = 0$
If $t_{1}, t_{2}, t_{3},$ and $t_{4}$ be the roots of this equation, then
$t_{1} + t_{2} + t_{3} + t_{4} = - (- \frac{4 m a^{2}}{m^{2} a^{2}}) = \frac{4}{m}$
$= 4 cot θ$
$cot θ = \frac{t_{1} + t_{2} + t_{3} + t_{4}}{4}$
$θ = {cot}^{- 1} (\frac{t_{1} + t_{2} + t_{3} + t_{4}}{4}) . . . . .1$
which is the required inclination
Again let $y_{1}$ and $y_{2}$ be the distances of the centroid from the axes of the first parabola; then $y_{1} = \frac{2 a (t_{1} + t_{2} + t_{3} + t_{4})}{4}$
$= 2 a cot θ$
Similarly, if $^{'} 4 b^{'}$ be the latus rectum of the second parabola, then
$y_{2} = 2 b cos θ$ , so $\frac{y_{1}}{y_{2}} = \frac{2 a cot θ}{2 b cot θ} = \frac{a}{b}$

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