Prove that the semi-latus-rectum of a parabola is a harmonic mean between the segments of any focal chord.

Question

Vedantu Content Team · Accepted Answer

Hint: t: We must assume a parabola, and find the length of its semi latus rectum. Then, we should assume 2 points on the parabola in parametric form, such that it is a focal chord. Keeping in mind the condition $t_{1} t_{2} = - 1$ for focal chords, and using the formula for harmonic mean $HM = \frac{2 a b}{a + b}$ , we can prove that this harmonic mean is equal to length of semi latus rectum.

Complete step by step answer:

Let us assume a parabola $y^{2} = 4 a x$ . We all know very well that a focal chord parallel to the directrix is called a latus rectum.

Also, for the parabola $y^{2} = 4 a x$ , the length of latus rectum is 4a. So, the length of the semi latus rectum will be 2a.

Let us assume a focal chord PQ and let point S be the focus of this parabola.

Since, the parabola is $y^{2} = 4 a x$ , we can say that the focus is S (a, 0).

Also, we know that the points P and Q lie on the parabola. SO, we can assume the points to be $P (a {t_{1}}^{2}, 2 a t_{1})$ and $Q (a {t_{2}}^{2}, 2 a t_{2})$ .

We need to prove that the harmonic mean of PS and QS will be equal to the length of semi latus rectum, i.e., 2a.

Using distance formula, we can write

$P S = \sqrt{{(a {t_{1}}^{2} - a)}^{2} + {(2 a t_{1})}^{2}}$

On simplifying, we can write

$P S = \sqrt{a^{2} {t_{1}}^{4} - 2 a^{2} {t_{1}}^{2} + a^{2} + 4 a^{2} {t_{1}}^{2}}$

$\Rightarrow P S = \sqrt{a^{2} {t_{1}}^{4} + 2 a^{2} {t_{1}}^{2} + a^{2}}$

$\Rightarrow P S = \sqrt{{(a {t_{1}}^{2} + a)}^{2}}$

$\Rightarrow P S = a ({t_{1}}^{2} + 1)$

Similarly, we can also write $Q S = a ({t_{2}}^{2} + 1)$ .

We know that the harmonic mean of two numbers, a and b, is defined as

$Harmonic mean = \frac{2 a b}{a + b}$ .

Thus, the harmonic mean of PS and QS, is

$H M = \frac{2 (P S) (Q S)}{(P S) + (Q S)}$

$\Rightarrow H M = \frac{2 a ({t_{1}}^{2} + 1) a ({t_{2}}^{2} + 1)}{a ({t_{1}}^{2} + 1) + a ({t_{2}}^{2} + 1)}$

We can simplify the above equation as

$H M = \frac{2 a ({t_{1}}^{2} {t_{2}}^{2} + {t_{1}}^{2} + {t_{2}}^{2} + 1)}{{t_{1}}^{2} + {t_{2}}^{2} + 2}$

Also, we know that for a focal chord, $t_{1} t_{2} = - 1$ . So, we get

$H M = \frac{2 a (1 + {t_{1}}^{2} + {t_{2}}^{2} + 1)}{{t_{1}}^{2} + {t_{2}}^{2} + 2}$

$\Rightarrow H M = \frac{2 a ({t_{1}}^{2} + {t_{2}}^{2} + 2)}{{t_{1}}^{2} + {t_{2}}^{2} + 2}$

Hence, the harmonic mean is equal to 2a.

Thus, the harmonic mean between the segments of any focal chord is equal to the semi latus rectum.

Note: We must understand that semi latus rectum is also a focal chord. Also, some students think that the harmonic mean of two numbers is the reciprocal of their arithmetic mean, which is not correct. We must not make such a mistake.