Question

Semi latus rectum of the parabola $y^2=4ax$, is the _____ mean between segments of any focal chord of the parabola.

• A. Arithmetic
• B. Geometric
• C. Harmonic
• D. None of these

Answer 1

The correct option is C

Harmonic

Standard parabola

$y^2=4ax$ is given

Length of latus rectum =4a

Semi latus rectum=2a

Let the coordinates of P be $(a{t}_1^2,2a{t}_1)\left)andQ\right((a{t}_2^2,2a{t}_2)$

$PS=\sqrt{(a-a{t}_1^2{)}^2+(2a{t}_1-0{)}^2}=a(1-t_1^2{)}^2+4a{t}_1^2$

$PS=\sqrt{a^2(1+t_1^2{)}^2}=a(1+t_1^2)$

$SQ=\sqrt{(a-a{t}_2^2{)}^2+(2a{t}_2-0{)}^2}=\sqrt{a(1-t_2^2{)}^2+4a{t}_2^2}$

$=\sqrt{a^2(1+t_2^2{)}^2}=a(1+t_2^2)$

Segment of the focul chord SP $\&$ SQ.

We need tha AM,GM and HM of these segments

Let's calculate HM

$HM=\frac{2\left(SP\right)\left(SQ\right)}{SP+SQ}=\frac{2a(1+t_1^2)\times a(1+t_2^2)}{a(1+t_1^2)+a(1+t_2^2)}$

$=\frac{2a^2(1+t_1^2)(1+t_2^2)}{a(1+t_1^2+1+t_2^2)}$

$=\frac{2a(1+t_1^2)(1+t_2^2)}{2+t_1^2+t_2^2)}$ ............(1)

Here, equation of focal chord

$y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)$

$y=2a{t}_1=\frac{2a{t}_2-2a{t}_1}{a{t}_2^2-a{t}_1^2}(x-a{t}_1^2)$

this passes through focus (a,0)

$0-2a{t}_1=\frac{2(t_2-t_1)}{(t_2-t_1)(t_2+t_1)}(a-a{t}_1^2)$

$-2a{t}_1=\frac{2}{t_1+t_2}a(1-t_1^2)$

$-t_1=\frac{1-t_1^2}{t_1+t_2}$

$-t_1^2-t_1.t_2=1-t_1^2$

$t_2=\frac{-1}{t_1}$

Substituting $t_2=\frac{-1}{t_1}$ in equation (1)

$HM=\frac{2a(1+t_1^2)(1+\frac{1}{t_1^2})}{2+t_1^2+\frac{1}{t_1^2}}$

$=\frac{\frac{2a(1+t_1^2)(1+t_1^2)}{t_1^2}}{\frac{2t_1^2+t_1^4+1}{t_1^2}}$

$=\frac{2a(1+t_1^2{)}^2}{(t_1^2+1{)}^2}=2a=$ Semi latus rectum

So, semi latus rectum of the parabola $y^2=4ax$ is the harmonic

mean between segments of any focal chord of parabola.

Similarly, we can try for AM $\&$ GM. We will find that it won't satisfy

the given condition of the question.