**
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.
**

Given the curve described by the following function *y = (1/2)x ^{2} - 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

It is important to start this exercise by drawing a quick graph of the situation :

We are looking for the point *(x _{2} ; y_{2})* of the parabola for which the tangent line

*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)x ^{2} - 2x + 3*. Its derivative is

*
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

*(x _{2} ; y_{2})*. Then we can say:

Parabola: *y = (1/2)x ^{2} - 2x + 3*

Tg2:

For *x = x _{2}* and

Parabola: *y _{2} = (1/2)(x_{2})^{2} - 2x_{2} + 3*

Tg2:

Thus *(1/2).(x _{2})^{2} - 2.x_{2} + 3 = a.x_{2} + b*

We apply the same logic to the intersection of the parabola and the tangent line *Tg1* at *(x _{1} ; y_{1})*. We get:

*(1/2).(x _{1})^{2} - 2.x_{1} + 3 = d.x_{1} + 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) *x _{1} - 2 = -1/a*

(5)

(6)

(7)

(8)

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, x_{1}, x_{2}, 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') *x _{1} = (-1/a) + 2*

(5')

(7')

(8')

Now let us use the equations (4') et (5') to replace every *x _{1}* by

(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'')

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

Reminder:

Given the second degree equation (quadratic equation) x, can be calculated as follow: _{2} |

Let us get back to our exercise with the equation (8''):

We have now to compute the value of *x _{2}* and

(5') *x _{2} = a + 2*

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

*x _{2} = 0,366 + 2 = 2,366* or

But,

Then for

And for

Conclusion:

At the point *(x _{2} ; y_{2}) = (2,366 ; 1,067)*, the tangent line to the parabola is perpendicular to the tangent line that passes through

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 *x _{2}* and