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

Correct Answer: 72

Parabola provided:
\[ y^2 = 12x \] 

Length of focal chord formula: 
\[ \text{Length of focal chord} = 4a \csc^2 \theta = 15 \] 
From the parabola, \(4a = 12\). Therefore: \[ 12 \csc^2 \theta = 15 \] 

Solving for \(\csc^2 \theta\): 
\[ \csc^2 \theta = \frac{15}{12} = \frac{5}{4} \]
This implies: \[ \sin^2 \theta = \frac{4}{5} \] 
Using the 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 focal chord PQ is \(\tan \theta = 2\). 

Equation of the chord through focus \((3, 0)\): \[ y - 0 = 2(x - 3) \] 

Simplified equation: \[ y = 2x - 6 \implies 2x - y - 6 = 0 \] 
Perpendicular distance from the origin \((0, 0)\) is calculated using: \[ p = \frac{|2 \times 0 - 0 - 6|}{\sqrt{2^2 + (-1)^2}} = \frac{6}{\sqrt{5}} \] 

Calculation of \(10p^2\): 
\[ 10p^2 = 10 \left(\frac{6}{\sqrt{5}}\right)^2 = 10 \times \frac{36}{5} = 72 \]