The sum of the reciprocals of focal distances of a focal chord PQ of y2 = 4ax is

Question

The sum of the reciprocals of focal distances of a focal chord PQ of y2 = 4ax is (A)  (1/a) (B) a (C) 2a (D)  (1/2a)

Tardigrade Admin · Accepted Answer

Equation of parabola y2=4 ax Total length of focal chord =4 a and half length of focal chord =2 a Now, sum of the reciprocals of focal distances =(1/2 a)+(1/2 a)=(2/2 a)=(1/a)

Q. The sum of the reciprocals of focal distances of a focal chord $PQ$ of $y^{2} = 4 a x$ is

$\frac{1}{a}$

$a$

$2 a$

$\frac{1}{2 a}$

Equation of parabola $y^{2} = 4 a x$
Total length of focal chord $= 4 a$
and half length of focal chord $= 2 a$
Now, sum of the reciprocals of focal distances
$= \frac{1}{2 a} + \frac{1}{2 a} = \frac{2}{2 a} = \frac{1}{a}$

Q. The sum of the reciprocals of focal distances of a focal chord PQ of y2=4ax is

a1​

a

2a

2a1​

Equation of parabola y2=4ax Total length of focal chord =4a and half length of focal chord =2a Now, sum of the reciprocals of focal distances =2a1​+2a1​=2a2​=a1​

Q. The sum of the reciprocals of focal distances of a focal chord $PQ$ of $y^{2} = 4 a x$ is

$\frac{1}{a}$

$a$

$2 a$

$\frac{1}{2 a}$

Equation of parabola $y^{2} = 4 a x$
Total length of focal chord $= 4 a$
and half length of focal chord $= 2 a$
Now, sum of the reciprocals of focal distances
$= \frac{1}{2 a} + \frac{1}{2 a} = \frac{2}{2 a} = \frac{1}{a}$