Well, I Guess That Works

Solution 1: Draw horizontal lines at both the points. The lines intersect the $y$ axis at $y = a^{2}$ and $y = b^{2}$ . Let's name the points $(0, a^{2} )$ and $( 0, b^{2} )$ $A$ and $B$ respectively. Let the string intersect the $y$ axis at $( 0, c )$ and this point is named $C$ . Then $C$ divides the line segment $AB$ in the ratio $a : - b$ .

Using the section formula , we can find the coordinates of $C$ .

$c = \frac{ a \times b^{2} +( - b) \times a^{2} }{a + (- b) } = \frac{ (a \times b ) (a - b ) } { a - b } = \boxed { a \times b } ~~~~~ _\square$

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Solution 2: Let's find the equation of the line formed by the string. Since we are know the coordinates of two points on the line, we can use the point-point form . The two points on the line are $( a, a^{2} )$ and $(-b, (-b)^{2} )$ .

The slope of the line is $m = \frac{ a^{2} - b^{ 2} } { a - ( -b ) } = \frac{ a^{2} - b^{ 2} } { a + b } = a - b$ . Hence the equation of the line is

$\begin{aligned} ( y - y_1) &= m ( x - x_1)\\ (y - a^2) & = (a - b ) \times ( x - a ) \end{aligned}$

To find the $y$ intercept, we set $x = 0$ .

$\begin{aligned} y - a^2 &= (a-b) \times (-a ) \\ y - a^{2} &= -a^{2} + a \times b \\ y &= \boxed{ a \times b } & _\square\\ \end{aligned}$

Looking at the figure,

When a=0, answer has to be 0. When b=0, answer has to be 0.

When a is fixed & b is moving (or vice-versa), y intercept is proportional to a (or b). Which makes the answer ab .

If the intercept is $(0,q)$ , we want $\frac{a^2-q}{a-0}=\frac{a^2-b^2}{a-(-b)}=a-b$ so $a^2-q=a^2-ab$ and $q=\boxed{ab}$ .