Question

Let the length of the focal chord $PQ$ of the parabola $y^2 = 12x$ be 15 units. If the distance of $PQ$ from the origin is $p$, then $10p^2$ is equal to _____

Answer 1

Given the parabola:
$y^2 = 12x$

The length of a focal chord is given by:
$\text{Length of focal chord} = 4a \csc^2 \theta = 15$
For this parabola, $4a = 12$, so: $12 \csc^2 \theta = 15$

Solving for $\csc^2 \theta$:
$\csc^2 \theta = \frac{15}{12} = \frac{5}{4}$
Thus: $\sin^2 \theta = \frac{4}{5}$
Using the trigonometric identity $\tan^2 \theta = \frac{\sin^2 \theta}{1 - \sin^2 \theta}$:
$\tan^2 \theta = \frac{\frac{4}{5}}{1 - \frac{4}{5}} = 4 \implies \tan \theta = 2$
The slope of the focal chord PQ is $\tan \theta = 2$.

The equation of the chord passing through the focus $(3, 0)$ is given by: $y - 0 = 2(x - 3)$

Simplifying: $y = 2x - 6 \implies 2x - y - 6 = 0$
To find the perpendicular distance of this line from the origin $(0, 0)$, use the formula:
$p = \frac{|2 \times 0 - 0 - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{6}{\sqrt{5}}$

Calculating $10p^2$:
$10p^2 = 10 \left(\frac{6}{\sqrt{5}}\right)^2 = 10 \times \frac{36}{5} = 72$