#### Mathematics online > Derivatives > Basic examples > Solved exercises > Intermediate level > Solution 8

Math derivative exercise : How figure out the X, Y coordinates of the point where the tangent line is perpendicular to the straight line determined by a given first degree equation ? Keywords: parabola equation, second degree equation, straight line equation, slope of a line, y-intercept, x-intercept.

### Derivatives : solution exercise 8 (Intermediate math level)

Given the curve described by the following function y = (1/2)x2 - 2x + 3, at which point of the graph the tangent line to the curve is perpendicular to another tangent line that passes through the point (1 ; 0) ?

It is important to start this exercise by drawing a quick graph of the situation : We are looking for the point (x2 ; y2) of the parabola for which the tangent line Tg2 is perpendicular to another tangent line Tg1 that passes through the point (1 ; 0). We know that the general cartesian equation of a straight line is y = ax + b :

Tg2:     y = ax + b
Tg1:     y = dx + c

Because Tg2 must be perpendicular to Tg1, we know that the product of their slope is equal to -1 :

a . d = -1

Which yields d = -1/a

The parabola is determined by the quadratic equation f(x) = (1/2)x2 - 2x + 3. Its derivative is
f '(x) = x - 2. We also know that f '(x1) = x1 - 2 = slope of Tg1 = d = -1/a. Thus
f '(x2) = x2 - 2 = slope of Tg2 = a. We also know that Tg1 passes through the point
(1 ; 0). Then we can say :

y = dx + c
0 = d.1 + c

Thus c = -d

We also know that Tg2 and the parabola share a same point whose coordinates are
(x2 ; y2). Then we can say:

Parabola: y = (1/2)x2 - 2x + 3
Tg2: y = ax + b

For x = x2 and y = y2 we have:

Parabola: y2 = (1/2)(x2)2 - 2x2 + 3
Tg2: y2 = ax2 + b

Thus (1/2).(x2)2 - 2.x2 + 3 = a.x2 + b

We apply the same logic to the intersection of the parabola and the tangent line Tg1 at (x1 ; y1). We get:

(1/2).(x1)2 - 2.x1 + 3 = d.x1 + c

Here is a short resumen of the equations we just found:

(1) Tg2:     y = ax + b
(2) Tg1:     y = dx + c
(3) d = - 1/a
(4) x1 - 2 = -1/a
(5) x2 - 2 = a
(6) c = - d
(7) (1/2).(x2)2 - 2.x2 + 3 = a.x2 + b
(8) (1/2).(x1)2 - 2.x1 + 3 = d.x1 + c

We have here a wonderful system of 8 equations to be solved. Before you start solving this system of equations, you must check that you have no more unknown variables than equations. If you have more unknown variables than equations in your system, you will not be able to solve it. Let us count how many unknown variables we have in this system of equations: y, a, x, b, d, c, x1, x2, which is a total of 8 unknown variables and 8 equations, it is all right then! We can now start to solve this system of 8 equations.

Let us take the equations (6) and (3) to replace every d by -1/a and every c by 1/a:

(1') Tg2:     y = ax + b
(2') Tg1:     y = -(1/a).x + 1/a
(4') x1 = (-1/a) + 2
(5') x2 = a + 2
(7') (1/2).(x2)2 - 2.x2 + 3 = a.x2 + b
(8') (1/2).(x1)2 - 2.x1 + 3 = (-1/a).x1 + 1/a

Now let us use the equations (4') et (5') to replace every x1 by (-1/a) + 2 and x2 by a + 2:

(1'') Tg2:     y = ax + b
(2'') Tg1:     y = (-1/a).x + 1/a
(7'') (1/2).(a + 2)2 - 2.(a + 2) + 3 = a.(a + 2) + b
(8'') (1/2).((-1/a) + 2)2 - 2.((-1/a) + 2) + 3 = (-1/a).((-1/a) + 2) + 1/a

The equation (8'') enables us to find a.

Reminder:

 Given the second degree equation (quadratic equation) If y = 0 we have In that case, the value of x, x1 and x2, can be calculated as follow: Let us get back to our exercise with the equation (8''): We have now to compute the value of x2 and y2 :

(5') x2 = a + 2

Two solutions have to be considered, one with a = 0,366 and the other with a = -1,366

x2 = 0,366 + 2 = 2,366 or x2 = -1,366 + 2 = 0,634
But, y2 = f(x2) = (1/2).(x2)2 - 2x2 + 3
Then for x2 = 2,366 we will have y2 = (1/2) . 2,3662 - 2 . 2,366 + 3 = 1,067
And for x2 = 0,634 we will have y2 = (1/2) . (0,634)2 - 2 . (0,634) + 3 = 1,933

Conclusion:

At the point (x2 ; y2) = (2,366 ; 1,067), the tangent line to the parabola is perpendicular to the tangent line that passes through (1 ; 0). The same thing happens at the point (0,634 ; 1,933) on the parabola. There are then two different solutions.

Feel free to use our free online Graph Plotter to trace the graph of the parabola online. Print that graph and you will be able to trace both tangent lines that pass through the point (1 ; 0) as well as both tangent lines perpendicular to the first one. All this to check graphically the values of x2 and y2 calculated.