On the curve y = x²-1 the points A (0,y_1) and B (1,y_2)

On the curve $y = x^{2}-1$ the points $A\bigl(0,y_{1}\bigr)$ and $B\bigl(1,y_{2}\bigr)$ are given. Using Lagrange's theorem, find the point $M$ of the arc $AB$ for which the area of the triangle $AMB$ is greatest, and compute that area.

Step 1: The two endpoints

\[ y_{1} = f(0) = -1, \qquad y_{2} = f(1) = 0, \qquad A = \bigl(0,-1\bigr),\ B = \bigl(1,0\bigr). \]

The chord has slope $\dfrac{0-(-1)}{1-0} = 1$ and equation $y = x-1$, i.e. $x-y-1 = 0$, with length $\bigl|AB\bigr| = \sqrt2$.

Step 2: The area as a function of the position of $M$

With $M = \bigl(\xi,\xi^{2}-1\bigr)$ and $\xi\in(0,1)$, the distance from $M$ to the chord is

\[ h(\xi) = \frac{\bigl|\xi-\bigl(\xi^{2}-1\bigr)-1\bigr|}{\sqrt{2}} = \frac{\xi\bigl(1-\xi\bigr)}{\sqrt2}, \]

the absolute value being unnecessary since $0\lt \xi\lt 1$. Hence

\[ S(\xi) = \frac12\bigl|AB\bigr|\,h(\xi) = \frac12\sqrt2\cdot\frac{\xi\bigl(1-\xi\bigr)}{\sqrt2} = \frac{\xi\bigl(1-\xi\bigr)}{2}. \tag{1} \]

Step 3: The mean value point

Lagrange's theorem applied to $f(x) = x^{2}-1$ on $[0,1]$ gives

\[ f'(\xi) = \frac{f(1)-f(0)}{1-0} = 1 \;\Longrightarrow\; 2\xi = 1 \;\Longrightarrow\; \xi = \frac12, \]

the point at which the tangent is parallel to the chord.

Step 4: Why that point maximises the area

Since the base $\bigl|AB\bigr|$ is fixed, maximising the area means maximising the distance to the chord. At the farthest point the tangent must be parallel to the chord — otherwise moving along the curve would increase the distance. Analytically, from (1),

\[ S'(\xi) = \frac{1-2\xi}{2} = 0 \;\Longrightarrow\; \xi = \frac12, \qquad S''= -1 \lt 0, \]

confirming a maximum at precisely the Lagrange point.

Step 5: Verification

\[S = \frac12\Bigl|x_{A}\bigl(y_{M}-y_{B}\bigr)+x_{M}\bigl(y_{B}-y_{A}\bigr)+x_{B}\bigl(y_{A}-y_{M}\bigr)\Bigr|\]

\[= \frac12\left|0+\frac12\cdot1+1\cdot\left(-\frac14\right)\right| = \frac12\cdot\frac14 = \frac18 \;\;\checkmark\]

\[\text{Tangent at } M:\ y = x-\tfrac54, \text{ parallel to the chord } y = x-1 \;\;\checkmark\]