Find the point where the tangent to the curve a/(x²)+
Find the point where the tangent to the curve $\dfrac{a}{x^2}+\dfrac{y}{b^2}=1$ meets the $x$-axis, as a function of the abscissa $x_0$ of the point of tangency.
Explanation of Method
Working the algebra through exactly shows the intercept is a cubic in x₀, not a pure proportionality.
Step 1: Solve for $y$ and differentiate
\[ y = b^2\!\left(1-\frac{a}{x^2}\right), \qquad y' = \frac{2ab^2}{x^3}. \tag{1} \]
At the point of tangency $(x_0,y_0)$: \[y_0 = b^2\!\left(1-\frac{a}{x_0^2}\right) = \frac{b^2(x_0^2-a)}{x_0^2}, \qquad y'(x_0) = \frac{2ab^2}{x_0^3}. \tag{2}\]
Step 2: The tangent line and its $x$-intercept
The tangent is $y-y_0=y'(x_0)(x-x_0)$; setting $y=0$ and solving for $x$, \[x_T = x_0 - \frac{y_0}{y'(x_0)}. \tag{3}\]
Substituting (2), \[\frac{y_0}{y'(x_0)} = \frac{\dfrac{b^2(x_0^2-a)}{x_0^2}}{\dfrac{2ab^2}{x_0^3}} = \frac{x_0(x_0^2-a)}{2a}, \tag{4}\]
\[ x_T = x_0 - \frac{x_0(x_0^2-a)}{2a} = x_0\cdot\frac{2a-(x_0^2-a)}{2a} = \frac{x_0(3a-x_0^2)}{2a}. \tag{5} \]
Step 3: Verification
\[a=2,\ b=1,\ x_0=3\text{: symbolic }x_T=\frac{3(6-9)}{4}=-2.25\text{, matching the direct tangent-line computation exactly}\ \checkmark\]
Result
Final answer:
\[\boxed{\,x_T = \frac{x_0(3a-x_0^2)}{2a}\,}\]
Remark
Remark. The intercept (5) is a genuine cubic polynomial in $x_0$ — it has both an $x_0$ term and an $x_0^3$ term — so $x_T$ is not proportional to $x_0^3$ for this curve; it only behaves like $-x_0^3/(2a)$ in the limit of large $x_0$, where the linear term becomes negligible. A pure proportionality $x_T\propto x_0^3$ would require the $3ax_0$ term in the numerator of (5) to vanish identically, which it does not for any fixed $a\neq0$. The exact closed form (5) is the honest statement of what the algebra gives for the curve as posed.