1. ## Addition of Intercepts for curve

Hi

I need help with this question:
a curve
square root(x) + Square root (y) = 1(eqn 1), for x and y bigger and equal too zero

Show that the sum of x any y intercepts of the tangent line to any point (a,b) on the curve (a,b bigger than 0) is equal to one

This is what i have done so far

Used implicit differentiation to find dy/dx of the eqn 1
It gives
dy/dx=-sqrt(y)/sqrt(x)

from there used y=mx+c

So m=-sqrt(y)/sqrt(x)
an then i subbed (a,b) into dy/dx to find m

so
that is y=(-sqrt(b)/sqrt(a))x+ c

then i subbed a(a,b) into y=(-sqrt(b)/sqrt(a))x+ c to find c
i found c= b+sqrt(ab)

from there i tried to find the x-intercept(used y=0) an y-intercept(used x=0)

so i found that x-int=sqrt(ab)-a
and y-int=b+sqrt(ab)

from there i've tried to relate it to original equation somehow but cannot relate it so that the sum of the intercepts equals 1

Any help would be appreciated, thank you

2. Originally Posted by Kevinsalborn
Hi

I need help with this question:
a curve
square root(x) + Square root (y) = 1(eqn 1), for x and y bigger and equal too zero

Show that the sum of x any y intercepts of the tangent line to any point (a,b) on the curve (a,b bigger than 0) is equal to one

This is what i have done so far

Used implicit differentiation to find dy/dx of the eqn 1
It gives
dy/dx=-sqrt(y)/sqrt(x)

from there used y=mx+c

So m=-sqrt(y)/sqrt(x)
an then i subbed (a,b) into dy/dx to find m

so
that is y=(-sqrt(b)/sqrt(a))x+ c
A simpler form is $y= m(x- x_0)+ y_0$ giving $y= \frac{-\sqrt{b}}{\sqrt{a}}(x- a)+ b$. That immediately gives $y= \frac{\sqrt{a}}{\sqrt{b}}x+ \sqrt{ab}+ b$ as you have.

then i subbed a(a,b) into y=(-sqrt(b)/sqrt(a))x+ c to find c
i found c= b+sqrt(ab)

from there i tried to find the x-intercept(used y=0) an y-intercept(used x=0)

so i found that x-int=sqrt(ab)-a
and y-int=b+sqrt(ab)

from there i've tried to relate it to original equation somehow but cannot relate it so that the sum of the intercepts equals 1

Any help would be appreciated, thank you
Very good!
With x= 0, $y= -\frac{\sqrt{b}}{\sqrt{a}}(-a)+ b= b+ \sqrt{ab}$ and with y= 0, $0= -\frac{\sqrt{b}}{\sqrt{a}}x+ b+ \sqrt{ab}$ so $\frac{\sqrt{b}}{\sqrt{a}}x= b+ \sqrt{ab}$, $x= \frac{\sqrt{a}}{\sqrt{b}}b+ \frac{\sqrt{a}}{\sqrt{b}}\sqrt{ab}= a+ \sqrt{ab}$. That is not exactly what you got- notice the "+ a" in the y-intercept.

$x-int+ y-int= b+ \sqrt{ab}+ a+ \sqrt{b}= b+ 2\sqrt{a}\sqrt{b}+ a$ and you ought to be able to recognize that as a "perfect square": $(\sqrt{a}+ \sqrt{b})^2$!

Now, use the fact that (a, b) is on the given curve so $\sqrt{a}+ \sqrt{b}= 1$.

3. I see it now, it was a perfect square, thank you so much for that