Why Is the Shortest Tangent Intercept of an Ellipse $a+b$?

Assume $a,b>0$. The statement is true when the “intercept made on the axes” means the length of the line segment cut off between the two coordinate axes. A tangent’s segment length is the distance between its $x$- and $y$-intercepts. We will derive that length, prove its lower bound, and find when equality holds.

There is a possible wording ambiguity: the length of the segment is not the same as the sum of the absolute values of the two intercept coordinates. The latter has a different minimum, noted below.

The tangent and its intercepts

A standard parametrization of the ellipse

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$

is $(a\cos t,b\sin t)$. The tangent at this point is

$$\frac{x\cos t}{a}+\frac{y\sin t}{b}=1.$$

Setting $y=0$ gives the $x$-intercept $a/\cos t$, and setting $x=0$ gives the $y$-intercept $b/\sin t$. For a tangent in another quadrant, one or both intercepts may be negative; their absolute values determine the segment length. When either denominator is zero, the tangent is parallel to an axis and has no finite segment between the axes, so it cannot give the least finite length.

For a tangent with finite intercepts, the distance formula gives

$$L=\sqrt{\frac{a^2}{\cos^2 t}+\frac{b^2}{\sin^2 t}}.$$

The signs do not affect $L$, since the intercept coordinates are squared.

Proof that the length is at least $a+b$

Because $\cos^2 t+\sin^2 t=1$, the Cauchy–Schwarz inequality gives

$$\left(\frac{a^2}{\cos^2 t}+\frac{b^2}{\sin^2 t}\right) (\cos^2 t+\sin^2 t)\ge (a+b)^2.$$

The second factor on the left is $1$. Therefore

$$L^2\ge (a+b)^2.$$

Since $L$ and $a+b$ are positive, taking square roots proves $L\ge a+b$. This establishes the lower bound for every tangent with finite axis intercepts.

The bound is attained, so it is a minimum rather than just a lower bound. Equality in Cauchy–Schwarz holds when

$$\frac{\cos^2 t}{a}=\frac{\sin^2 t}{b}.$$

Together with $\cos^2 t+\sin^2 t=1$, this yields

$$\cos^2 t=\frac{a}{a+b},\qquad \sin^2 t=\frac{b}{a+b}.$$

These values are possible for a real parameter $t$. Substituting them into the length formula gives $L=a+b$. Thus the least length of the tangent segment between the axes is exactly $a+b$.

Worked example: semiaxes 4 and 3

For the ellipse $x^2/16+y^2/9=1$, the least tangent-segment length is

$$4+3=7.$$

One equality case has $\cos t=2/\sqrt{7}$ and $\sin t=\sqrt{3/7}$. Its intercepts are

$$x=\frac{4}{2/\sqrt{7}}=2\sqrt{7},\qquad y=\frac{3}{\sqrt{3/7}}=\sqrt{21}.$$

The distance between them is

$$\sqrt{(2\sqrt{7})^2+(\sqrt{21})^2} =\sqrt{28+21}=7.$$

The tangent is therefore one that actually achieves the bound, not merely one whose length approaches it.

Common mistakes and a wording distinction

• Adding the intercepts instead of measuring the segment: The distance between points $(p,0)$ and $(0,q)$ is $\sqrt{p^2+q^2}$, not $p+q$. Here $p$ and $q$ are the intercept coordinates.
• Ignoring signs: Some tangents have negative intercepts. Use absolute values for lengths, or square the coordinates in the distance formula.
• Treating an axis-parallel tangent as a finite candidate: Such a tangent does not cut off a finite segment between the axes.
• Confusing two optimization problems: If “least intercept” is meant to ask for the least sum of the absolute intercept coordinates, that sum is not generally $a+b$. Its minimum over tangents is $\bigl(a^{2/3}+b^{2/3}\bigr)^{3/2}$. The $a+b$ result concerns the Euclidean length of the segment cut off by the axes.