Prove that three tangents to a parabola, which are such that the tangents of their inclinations to the axis are in a given harmonical progression, form a triangle whose area is constant.

Open in App

Solution

$y^{2} = 4 a x$
eqn of tangent in parametric form at $(a t^{2}, 2 a t)$
$t y = x + a t^{2}$
three tangents to parabola are
$t_{1} y = x + a t_{1}^{2}$ .................(1)
$t_{2} y = x + a t_{2}^{2}$ .................(2)
$t_{3} y = x + a t_{3}^{2}$ .................(3)
given that slopes are in H.P
$- \frac{1}{t_{1}}$ , $- \frac{1}{t_{2}}$ , $- \frac{1}{t_{3}}$ are in H.P
or $- t_{1}, - t_{2}, - t_{3}$ are in A.P
$\Rightarrow 2 t_{2} = t_{1} + t_{3}$
area of triangle $= \frac{1}{2} .2 a^{2} ∣ ∣ t_{1} t_{2} (t_{1} - t_{2}) + t_{3}^{2} (t_{1} - t_{2}) + t_{3} (t_{1}^{2} - t_{2}^{2}) ∣ ∣$
area of triangle $= a^{2} ∣ ∣ t_{1} t_{2} (t_{1} - t_{2}) + t_{3}^{2} (t_{1} - t_{2}) + t_{3} (t_{1}^{2} - t_{2}^{2}) ∣ ∣$
$= c o n s t a n t$

Suggest Corrections

Similar questions

4 a^{2} s e c^{3} θ c o s e c^{3} θ

x = 2 a

α

y = (x - a) tan 2 α

x^{2} + y^{2} - 8 x + 12 = 0

Join BYJU'S Learning Program