The graphs of y = sqrt(x+a) and y = sqrt(2x-x^2) have the same gradient at their point of intersection. Find a and the point of intersection. (a is a constant)

Question

A full worked solution would be very helpful.

Answer 1

We have system of equations for x and a: y₁ = y₂ and y₁' = y₂';

We have: sqrt(x + a) = sqrt(2x - x²);

1/(2sqrt(x + a)) = (1 - x)/(sqrt(2x - x²));

Squering both sides of these equations we get: x + a = 2x - x²; and 4(1 - x)²(x + a) = 2x - x²,

4(1 - x)²(2x - x²) = 2x - x²; (2x - x²)[4(1 - x)² - 1] = 0;

From here x = 0, 2 - x = 0 or x = 2; 4(1 - x)² - 1 = 0; (x - 1)² = 1/4; x - 1 = ±1/2; x = 3/2 or x = 1/2

We get for x 4 values: 0, 2, 3/2, 1/2.

If x = 0 then sqrt(0+a) = sqrt(2·0 - 0²);p sqrta = sqrt0, a = 0. Now y-coordinate of popint of intersection is 0.

For x = 0 a = 0 and POI is (0,0), gradients vertical

For x = 2 sqrt(2 + a) = sqrt(2·2 - 2²) = sqrt0, 2 + a = 0 and a = - 2, gradients also vertical; POI (2, 0)

For x = 3/2 sqrt(3/2 + a) = sqrt(2·3/2 - 9/4); 3/2 + a = 3/4; a = - 3/4and second equation not satisfied.

For x = 1/2 sqrt(1/2 + a) = sqrt(1 - 1/4), 1/2 + a = 3/4 and a = 1/4, y = sqrt(3/4) and POI is (1/2, sqrt(3/4))

Answer: a = 0, (0,0); a = 2, (2,0); a = 1/4, (1/2, sqrt3/4)