# Given the quadratic `Q(x) = x^2 - 2x_0x + 4ay_0` suppose the points A, B lie on a line tangential to Q(x) and that the x-coordinates of A and B are equal to the x-coordinates of the roots of...

`Q(x) = x^2 - 2x_0x + 4ay_0`

suppose the points A, B lie on a line tangential to Q(x) and that the x-coordinates of A and B are equal to the x-coordinates of the roots of Q(x).

Find the coordinates of the points A and B given that the line tangential to Q(x) touches Q(x) at `x=x_1`.

mathsworkmusic | (Level 2) Educator

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First find the x-coordinates of A and B respectively, which are equal to the two roots of the quadratic Q(x). To find these roots `x(A)``x(B)` we solve

`Q(x) = 0`

for `x`

Using the quadratic formula `x = (-beta pm sqrt(beta^2-4alpha gamma))/(2 alpha)`  where we write

`Q(x)` as  `alpha x^2 + beta x + gamma`   we then have that

`x(A)``x(B)` `= (2x_0 pm sqrt(4x_0^2 - 16ay_0))/2`  `= x_0 pm sqrt(x_0^2 - 4ay_0)`

Now, the tangent line to Q(x) on which A and B lie has gradient equal to the gradient function of Q(x), Q'(x). We obtain this gradient function by differentiating Q(x) with respect to x:

`Q'(x) = 2(x - x_0)`

Since the tangent line touches Q(x) at (x,Q(x)) we know that that point is on the tangent line.

Consider a point `x_1` where a particular tangent line `T_1(x)` touches the parabola Q(x). ``The gradient of `T_1(x)` is the gradient of Q(x) at `x_1``2(x_1-x_0)``T_1(x)` also passes through the point (`x_1, Q(x_1))` so that` `

`T_1(x) = 2(x_1-x_0)x + Q(x_1) - 2(x_1-x_0)x_1`

`= 2(x_1-x_0)(x-x_1) + Q(x_1)`.

Since this is the equation for the tangent line on which A and B lie then the y-coordinates of A and B are given by `T_1(x(A))`  and `T_1(x(B))` respectively.

Note that if  `x_0^2 < 4ay_0` (the discriminant of Q(x) is < 0 ), then A and B do not exist as the parabola doesn't cross the x-axis at all. If  `x_0^2 = 4ay_0` then A=B as the parabola touches the x-axis at A=B.