Find the equation of the tangent in the given problem

[chwala]

Homework Statement

See attached

Relevant Equations

Differentiation

The textbook solution is indicated below;

My approach on this question is as follows;

$\dfrac{dy}{dx} = -\dfrac {8}{x^3}$

$\dfrac{dy}{dx} (x=a) = -\dfrac {8}{a^3}$

The tangent equation is given by;

$y= -\dfrac {8}{a^3}x+\dfrac{12+a^2}{a^2}$

when $x=0$,

$y=\dfrac{12+a^2}{a^2}$

and when $y=0$,

$\dfrac {8}{a^3} x = \dfrac{12+a^2}{a^2}$

$⇒\dfrac {8}{a^2}=y$

$⇒\dfrac {8}{a^2}=\dfrac{4}{a^2} +1$

$8=4+a^2$

$4=a^2$

$a=\sqrt{4}=2$

on using; $y=\dfrac{12+a^2}{a^2}$ and substituting $a=2$ yields

$y=\dfrac{12+2^2}{2^2}=\dfrac{16}{4}=4$

$b=4$.

I would appreciate any other method or insight. Cheers guys!

 

[fresh_42]

What happened to $a=-2$ at $y=0$?

I think your method is fine, no need to make things more complicated.

 

[topsquark]

Homework Statement:: See attached
Relevant Equations:: Differentiation

View attachment 316098The textbook solution is indicated below;

View attachment 316099

My approach on this question is as follows;

$\dfrac{dy}{dx} = -\dfrac {8}{x^3}$

$\dfrac{dy}{dx} (x=a) = -\dfrac {8}{a^3}$

The tangent equation is given by;

$y= -\dfrac {8}{a^3}x+\dfrac{12+a^2}{a^2}$

when $x=0$,

$y=\dfrac{12+a^2}{a^2}$

and when $y=0$,

$\dfrac {8}{a^3} x = \dfrac{12+a^2}{a^2}$

$⇒\dfrac {8}{a^2}=y$

$⇒\dfrac {8}{a^2}=\dfrac{4}{a^2} +1$

$8=4+a^2$

$4=a^2$

$a=\sqrt{4}=2$

on using; $y=\dfrac{12+a^2}{a^2}$ and substituting $a=2$ yields

$y=\dfrac{12+2^2}{2^2}=\dfrac{16}{4}=4$

$b=4$.

I would appreciate any other method or insight. Cheers guys!

My only thought is that, once you get the slope, the line that passes through the point (0, b) is $(y - b) = m(x - 0 )$, and the line through (b, 0) is $(y - 0) = m(x - b)$. You can solve those simultaneously. However I don't think that approach gets you anything that much simpler.

-Dan

 

[chwala]

What happened to $a=-2$ at $y=0$?

I think your method is fine, no need to make things more complicated.

You are right; $a=±2$, We shall treat $a=-2$ as unsuitable as it will not realize the required co-ordinates at the $x$ axes, that is $(b,0)$ and $y$ axes, that is $(0,b)$.

 

[pasmith]

The conditions on the tangent line intersecting the axes require that its equation be $$f(a) + f'(a)(x - a) = \frac{12}{a^2} + 1 - \frac{8}{a^3}x = y = b - x.$$ Comparing powers of $x$ gives $$\begin{split}<br /> \frac{12}{a^2} + 1 &= b \\<br /> \frac{8}{a^3} &= 1.\end{split}$$ These are easily solved to find $a = 2$ and $b = 4$.