Tangent and f(x) = sqrt[ax - x^2], a>0 constant: distance of origin from....

[orges13]

exercise

H

 

[Otis]

Hello guys,
I am struggling with this question.

---

Prove that the distance of the origin O from the tangent at a point P of the graph of the following function:

. . . . .\(\displaystyle f(x)\, =\, \sqrt{\strut ax\, -\, x^2\,},\, \mbox{ where }\, a\, >\, 0\, \mbox{ is constant}\)

...is equal to the distance of the point P from the y-axis.

---

... i built the figure ...

Something seems wrong with the wording of this exercise, and I don't understand why it includes this phrase: "the tangent at".

The distance of an arbitrary point P from the y-axis is always x.

But the distance of the Origin from any point (on the graph) equals x at only one location: (a,0)

In other words, the two distances are equal when x=a, and they are not equal at any other value in the domain.

So this exercise seems to ask for a proof of something that is false for all possible points P except (a,0).

Also, what is the meaning of "built the figure"?

Are you saying that you picked some values for a and plotted the respective graphs?

If so, then you saw that f's graph is the upper-half of a circle with radius a/2, centered at the point (a/2,0). This may also be seen algebraically.

If any point on this semi-circle is to be the same distance from the y-axis as it is from the origin, then it must be the point (a,0).

Additionally, the tangent line at (a,0) has undefined slope because f'(x) is not defined at x=a.