If the x-intercept of a focal chord of the parabola y²=8x+4y+4 is 3, then the length of this chord is equal to ____________.

Solution

Maths sir · 5 steps

Complete the square to find the focus, then the focus and (3, 0) fix the chord's slope.

1. Standard form. Move 4y across and complete the square: y² - 4y + 4 = 8x + 8, so (y-2)² = 8(x+1). This is Y² = 4aX with X = x+1, Y = y-2 and a = 2.
2. Find the focus. The vertex is (-1, 2). The focus is a = 2 units to its right, at (1, 2).
3. Slope of chord. The chord passes through the focus (1, 2) and its x-intercept (3, 0). Its slope is m = (0-2)/(3-1) = -1, so m² = 1.

   L = 4a(1+m²)/m² = 4a cosec²θ

4. Put in numbers. L = (4 × 2 × (1+1))/1 = 16. As a check, put X = 2 - Y into Y² = 8X: the roots are Y = -4 ± 4√2, they differ by 8√2, and the length is √2 × 8√2 = 16.
5. So. The focal chord has length 16.

The idea to keep. A shifted parabola is still Y² = 4aX after you complete the square. Find the real focus first, and a focal chord with slope m has length 4a(1+m²)/m².

Answer from the NTA final answer key. Solution by PrepWiser.