Prove that the semi-latus-rectum of a parabola is a harmonic mean between the segments of any focal chord.
VerifiedHint: 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 $\text{HM}=\dfrac{2ab}{a+b}$, 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 ${{y}^{2}}=4ax$. We all know very well that a focal chord parallel to the directrix is called a latus rectum.
Also, for the parabola ${{y}^{2}}=4ax$, 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 ${{y}^{2}}=4ax$, 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 $P\left( a{{t}_{1}}^{2},2a{{t}_{1}} \right)$ and $Q\left( a{{t}_{2}}^{2},2a{{t}_{2}} \right)$.
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
\[PS=\sqrt{{{\left( a{{t}_{1}}^{2}-a \right)}^{2}}+{{\left( 2a{{t}_{1}} \right)}^{2}}}\]
On simplifying, we can write
\[PS=\sqrt{{{a}^{2}}{{t}_{1}}^{4}-2{{a}^{2}}{{t}_{1}}^{2}+{{a}^{2}}+4{{a}^{2}}{{t}_{1}}^{2}}\]
\[\Rightarrow PS=\sqrt{{{a}^{2}}{{t}_{1}}^{4}+2{{a}^{2}}{{t}_{1}}^{2}+{{a}^{2}}}\]
\[\Rightarrow PS=\sqrt{{{\left( a{{t}_{1}}^{2}+a \right)}^{2}}}\]
\[\Rightarrow PS=a\left( {{t}_{1}}^{2}+1 \right)\]
Similarly, we can also write \[QS=a\left( {{t}_{2}}^{2}+1 \right)\].
We know that the harmonic mean of two numbers, a and b, is defined as
$\text{Harmonic mean}=\dfrac{2ab}{a+b}$.
Thus, the harmonic mean of PS and QS, is
$HM=\dfrac{2\left( PS \right)\left( QS \right)}{\left( PS \right)+\left( QS \right)}$
$\Rightarrow HM=\dfrac{2a\left( {{t}_{1}}^{2}+1 \right)a\left( {{t}_{2}}^{2}+1 \right)}{a\left( {{t}_{1}}^{2}+1 \right)+a\left( {{t}_{2}}^{2}+1 \right)}$
We can simplify the above equation as
$HM=\dfrac{2a\left( {{t}_{1}}^{2}{{t}_{2}}^{2}+{{t}_{1}}^{2}+{{t}_{2}}^{2}+1 \right)}{{{t}_{1}}^{2}+{{t}_{2}}^{2}+2}$
Also, we know that for a focal chord, ${{t}_{1}}{{t}_{2}}=-1$. So, we get
$HM=\dfrac{2a\left( 1+{{t}_{1}}^{2}+{{t}_{2}}^{2}+1 \right)}{{{t}_{1}}^{2}+{{t}_{2}}^{2}+2}$
$\Rightarrow HM=\dfrac{2a\left( {{t}_{1}}^{2}+{{t}_{2}}^{2}+2 \right)}{{{t}_{1}}^{2}+{{t}_{2}}^{2}+2}$
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.
