sum of intercepts of a tangent line
- Thread starter emilyf
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emilyf said:
A tangent line to the given curve will pass through the tangent point - which in this case is [x[sub:32gt193e]1[/sub:32gt193e],(?c - ?x[sub:32gt193e]1[/sub:32gt193e])[sup:32gt193e]2[/sup:32gt193e]]
What is the slope of the tangent line - (think about derivative of the function)?
What is the equation of the tangent line - you know the slope and a point on the line?
What are the intercepts?
Add those up and show....
Let's try it by using implicit differentiation.
\(\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{c}\)
\(\displaystyle y'=-\sqrt{\frac{y}{x}}\)
Now, let (a,b) be a point on the curve (I only chose (a,b) as an arbitrary point because they go with 'c' more than, say, (p,q)).
Thus, we get:
\(\displaystyle \sqrt{a}+\sqrt{b}=\sqrt{c}\)
\(\displaystyle y'=-\sqrt{\frac{b}{a}}\)
Now, using the line equation \(\displaystyle (y-y_{1})=m(x-x_{1})\) we get:
\(\displaystyle (y-b)=-\sqrt{\frac{b}{a}}(x-a)\)
If \(\displaystyle x=0\Rightarrow y=b+\sqrt{ab}\)
If \(\displaystyle y=0\Rightarrow x=a+\sqrt{ab}\)
Now, can you finish?. Almost there.
Add x and y, then factor and see what you get.
\(\displaystyle \sqrt{x}+\sqrt{y}=\sqrt{c}\)
\(\displaystyle y'=-\sqrt{\frac{y}{x}}\)
Now, let (a,b) be a point on the curve (I only chose (a,b) as an arbitrary point because they go with 'c' more than, say, (p,q)).
Thus, we get:
\(\displaystyle \sqrt{a}+\sqrt{b}=\sqrt{c}\)
\(\displaystyle y'=-\sqrt{\frac{b}{a}}\)
Now, using the line equation \(\displaystyle (y-y_{1})=m(x-x_{1})\) we get:
\(\displaystyle (y-b)=-\sqrt{\frac{b}{a}}(x-a)\)
If \(\displaystyle x=0\Rightarrow y=b+\sqrt{ab}\)
If \(\displaystyle y=0\Rightarrow x=a+\sqrt{ab}\)
Now, can you finish?. Almost there.
Add x and y, then factor and see what you get.