Let the quadratic curve passing through the point (-1,0) and touching the line y=x at (1,1) be y=f(x). Then the x-intercept of the normal to the curve at the point (α, α+1) in the first quadrant is __________.

Solution

Maths sir · 5 steps

Three conditions fix the three coefficients of the quadratic. Then find the point and draw the normal there.

1. Set up conditions. Take f(x) = ax² + bx + c. It passes through (-1,0), so a - b + c = 0. It touches y = x at (1,1), so f(1) = 1 and f'(1) = 1. That gives a + b + c = 1 and 2a + b = 1.
2. Solve for coefficients. Subtracting the first two equations gives 2b = 1, so b = 1/2. Then a = 1/4 and c = b - a = 1/4, so f(x) = (x+1)²/4.
3. Find the point. Put (α, α+1) on the curve: (α+1)²/4 = α + 1. This gives α = -1 or α = 3. Only α = 3 is in the first quadrant, so the point is (3, 4).
4. Slope of normal. f'(x) = (x+1)/2, so the tangent slope at x = 3 is 2. The normal slope is -1/2.

   y - 4 = -1/2(x - 3)

5. So. Put y = 0: x - 3 = 8, so x = 11. The x-intercept of the normal is 11.

The common slip: 1. Using the tangent slope 2 instead of the normal slope -1/2 gives the tangent's x-intercept, 1, not 11.

The idea to keep. Touching a line at a point gives two conditions: the curve passes through the point, and its slope there equals the slope of the line. For the normal, use the negative reciprocal of the tangent slope.

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