Author Topic: Advanced question (Read 612 times) Tweet Share

kenhung123 · « **on:** June 26, 2010, 02:13:18 pm »

I have absolutely no idea for this question seems too hard for me..

brightsky · « **Reply #1 on:** June 26, 2010, 03:56:01 pm »

From the above, we know that $f'(x_1) = m$ .

So we know that the gradient of the second tangent would be:

$abf'\left(b \cdot \frac{x_1}{b}\right) = abf'(x_1) = abm$ [Use of chain rule]

The equation of the tangent of y = f(x) at $(x_1, y_1)$ can also be expressed as:

$y - y_1 = m (x - x_1)$

$y = mx - mx_1 + y_1$

This means that $-mx_1 + y_1 = c$ ...(1)

Likewise, the equation of the second tangent can be expressed as:

$y - y_1a = abm\left(x - \frac{x_1}{b}\right)$

$y - y_1a = abmx - amx_1$

$y = abmx - amx_1 + y_1a$

$y = abmx + a(-mx_1 + y_1)$

$y = abmx + ac$ [Substitute (1) into equation]

$y = a(mbx + c)$

Q.E.D

kenhung123 · « **Reply #2 on:** June 26, 2010, 05:50:54 pm »

Damn your good. Thanks brightsky!

Hutchoo · « **Reply #3 on:** June 26, 2010, 09:08:45 pm »

^ He sure is.

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