Show that the semi latus rectum of the parabola y2 = 4ax is a harmonic mean between the segment of - Maths - Conic Sections

Question

Show that the semi latus rectum of the parabola y2 =
4ax is a harmonic mean between the segment of any focal chord

Lalit Mehra · Accepted Answer

Equation of the given parabola is y2 = 4ax Coordinates of focus = S(a, 0) Let $P (a t_{1}^{2}, 2 a t)$ and $Q (a t_{2}^{2}, 2 a t_{2})$ be the end point of the focal chord of the given parabola âˆ´ t1 t2 = â€“ 1 (1) Length of the semi latus rectum of the given parabola = 2a Let SP and SQ be the segment of the focal chord $∴ SP = a + x = a + a t_{1}^{2} = a (1 + t_{1}^{2})$ Similarly, $SQ = a (1 + t_{2}^{2}) = a [1 + (- \frac{1}{t_{1}})^{2}] = a (1 + \frac{1}{t_{1}^{2}}) = \frac{a (1 + t_{1}^{2})}{t_{1}^{2}} [U \sin g (1)] ∴ \frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a (1 + t_{1}^{2})} + \frac{t_{1}^{2}}{a (1 + t_{1}^{2})} = \frac{1 + t_{1}^{2}}{a (1 + t_{1}^{2})} = \frac{1}{a} \Rightarrow \frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a} = \frac{2}{2 a} \Rightarrow \frac{1}{SP}, \frac{1}{2 a} and \frac{1}{SQ} are in A P$ â‡’ SP, 2a and SQ are in H P Thus, the semi latus rectum of the given parabola is the harmonic mean between the segment of the local chord

Show that the semi latus rectum of the parabola y2 = 4ax is a harmonic mean between the segment of any focal chord.

Show that the semi latus rectum of the parabola y² = 4ax is a harmonic mean between the segment of any focal chord.