Find the equation of the tangent in the given problem

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Homework Statement
See attached
Relevant Equations
Differentiation
1666782546246.png
The textbook solution is indicated below;

1666782657824.png


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!
 
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What happened to ##a=-2## at ##y=0##?

I think your method is fine, no need to make things more complicated.
 
chwala said:
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
 
fresh_42 said:
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)##.
 
The conditions on the tangent line intersecting the axes require that its equation be <br /> f(a) + f&#039;(a)(x - a) = \frac{12}{a^2} + 1 - \frac{8}{a^3}x = y = b - x. Comparing powers of x gives <br /> \begin{split}<br /> \frac{12}{a^2} + 1 &amp;= b \\<br /> \frac{8}{a^3} &amp;= 1.\end{split} These are easily solved to find a = 2 and b = 4.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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