Consider the parabola with focus and directrix , as shown in the figure. Among the points on this parabola, let be a point in the first quadrant, and let be the foot of the perpendicular from to the directrix. Consider the ellipse whose foci lie on the -axis and which passes through the three points , , and . Let be the focus of this ellipse with positive -coordinate. Suppose that triangle has perimeter and area . Let the length of segment be . Compute . [4 points]
To solve the problem, we need to find ( k^2 ), where ( k ) is the length of segment ( \overline{HF} ) for the given parabola and ellipse conditions.
Step 1: Parabola Equation and Point Coordinates
The parabola has focus ( F(p, 0) ) and directrix ( x = -p ). The equation of the parabola is: [ y^2 = 4px ] Point ( A ) on the parabola in the first quadrant has coordinates ( (x_1, y_1) ), satisfying: [ y_1^2 = 4px_1 ] Point ( H ), the foot of the perpendicular from ( A ) to the directrix, has coordinates ( (-p, y_1) ).
Step 2: Length of ( \overline{HF} )
The length ( k ) of segment ( \overline{HF} ) is given by: [ k = \sqrt{(p - (-p))^2 + (0 - y_1)^2} = \sqrt{(2p)^2 + y_1^2} = \sqrt{4p^2 + y_1^2} ] Thus, [ k^2 = 4p^2 + y_1^2 ]
Step 3: Area and Perimeter Conditions
The area of triangle ( AHB ) is given by: [ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (x_1 + p) \times y_1 = 2p + 12 ] This simplifies to: [ (x_1 + p) y_1 = 4p + 24 ]
The perimeter of triangle ( AHB ) is given by: [ (x_1 + p) + \sqrt{(c + p)^2 + y_1^2} + \sqrt{(x_1 - c)^2 + y_1^2} = p + 27 ] Simplifying, we get: [ \sqrt{(c + p)^2 + y_1^2} + \sqrt{(x_1 - c)^2 + y_1^2} = 27 - x_1 ]
Step 4: Solving the System of Equations
We have the following system of equations:
- ( y_1^2 = 4px_1 )
- ( (x_1 + p) y_1 = 4p + 24 )
Substitute ( x_1 = \frac{y_1^2}{4p} ) into the second equation: [ \left( \frac{y_1^2}{4p} + p \right) y_1 = 4p + 24 ] Multiply through by ( 4p ): [ (y_1^2 + 4p^2) y_1 = 16p^2 + 96p ] This simplifies to: [ y_1^3 + 4p^2 y_1 - 16p^2 - 96p = 0 ]
Step 5: Finding Real Solutions
We solve the cubic equation ( y_1^3 + 4p^2 y_1 - 16p^2 - 96p = 0 ). By trial, we find that ( y_1 = 6 ) and ( p = 3 ) satisfy the equation: [ 6^3 + 4(3)^2(6) - 16(3)^2 - 96(3) = 216 + 216 - 144 - 288 = 0 ]
Step 6: Verifying the Solution
For ( p = 3 ) and ( y_1 = 6 ): [ x_1 = \frac{6^2}{4 \cdot 3} = 3 ] The perimeter condition is: [ \sqrt{(c + 3)^2 + 6^2} + \sqrt{(3 - c)^2 + 6^2} = 24 ] Solving this, we find ( c = \sqrt{\frac{528}{5}} ), which satisfies the equation.
Step 7: Calculating ( k^2 )
Finally, we calculate ( k^2 ): [ k^2 = 4p^2 + y_1^2 = 4(3)^2 + 6^2 = 36 + 36 = 72 ]
Thus, the value of ( k^2 ) is: [ \boxed{72} ]