Question

Let the tangent to the parabola \(y^2 = 12x\) at the point (3, α) be perpendicular to the line 2x + 2y = 3. Then the square of distance of the point (6, – 4) from the normal to the hyperbola \(α^2 x^2 – 9y^2 = 9α^2\) at its point (α– 1, α + 2) is equal to__

Answer 1

Correct Answer: 116

1. Given: The point P(3, α) lies on the parabola \( y^2 = 12x \). Substituting \( x = 3 \):  
   \( \alpha^2 = 12 \cdot 3 = 36 \) → \( \alpha = \pm6 \).
2. Determine the correct value of α: 
   The slope of the tangent at \( P(3, \alpha) \) is: 
   \( \frac{dy}{dx} = \frac{6}{\alpha} \). 
   Given that the slope is 1: 
   \( \frac{6}{\alpha} = 1 \) → \( \alpha = 6 \) (Reject \( \alpha = -6 \)).
3. Equation of the hyperbola: 
   The hyperbola is given as: 
   \( \frac{x^2}{9} - \frac{y^2}{36} = 1 \).
4. Point Q: 
   Point \( Q(\alpha - 1, \alpha + 2) = (3 - 1, 6 + 2) = (2, 8) \). 
   Substitute \( Q(2, 8) \) into the hyperbola equation: 
   \( \frac{9x}{5} + \frac{36y}{8} = 45 \).
5. Equation of the normal: 
   The normal passes through \( (3, 6) \): 
   \( 2x + 5y - 50 = 0 \).
6. Distance of point \( (6, -4) \) from the normal: 
   The distance of \( (6, -4) \) from the line \( 2x + 5y - 50 = 0 \) is: 
   \[ \text{Distance} = \frac{|2(6) + 5(-4) - 50|}{\sqrt{2^2 + 5^2}} \] Simplify: 
   \( \text{Distance} = \frac{|12 - 20 - 50|}{\sqrt{4 + 25}} = \frac{58}{\sqrt{29}} \).
7. Square of the distance: 
   The square of the distance is: 
   \( \left(\frac{58}{\sqrt{29}}\right)^2 = \frac{3364}{29} = 116 \).

Final Answer:

The square of the distance is 116.