Let a line perpendicular to the line 2 x-y=10 touch the parabola y²=4(x-9) at the point P. The distance of the point P from the centre of the circle x²+y²-14 x-8 y+56=0 is __________.

Solution

Maths sir · 5 steps

Shift the parabola to standard form, use the slope-point formula for P, then measure to the centre.

1. Slope of tangent. The line 2x - y = 10 has slope 2. A line perpendicular to it has slope m = -1/2.
2. Standard form. Put X = x - 9, Y = y. The parabola becomes Y² = 4X, so a = 1. A tangent of slope m to Y² = 4aX touches it at (a/m², 2a/m).

   (1/(-1/2)², 2/(-1/2)) = (4, -4)

3. Shift back. x = X + 9 = 13 and y = -4, so P = (13, -4). As a check, y² = 16 = 4(13 - 9), and the slope there is 2/y = -1/2.
4. Centre of circle. For x² + y² - 14x - 8y + 56 = 0 the centre is (7, 4). The distance is √((13-7)² + (-4-4)²) = √(36 + 64) = √100.
5. So. The distance of P from the centre is 10.

The common slip: 3.75. Using slope 2 instead of the perpendicular slope -1/2 puts P at (9.25, 1) and gives 3.75, not 10.

The idea to keep. For a shifted parabola, move it to Y² = 4aX first. Get the point of contact from (a/m², 2a/m), then shift back before you measure any distance.

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