Prove that the least intercept made on the tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ by the axes is $a+b$.

5,115

Solution 1

The end points are, for each $\theta$, $$X\left(\frac a{\cos\theta},0\right)\qquad Y\left(0,\frac b{\sin\theta}\right)$$

Then define $$f(\theta)=d(X,Y)=\frac1{|\sin\theta\cos\theta|}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}$$

Since this not depend on the quadrant where the contact point is, we can assume WLOG that $\theta$ is an acute angle. Define now

$$g(\theta)=\ln f(\theta)=-\ln\sin\theta-\ln\cos\theta+\frac12\ln(a^2\sin^2\theta+b^2\cos^2\theta)$$

and differentiate:

$$g'(\theta)=-\cot\theta+\tan\theta+\frac{(a^2-b^2)\sin\theta\cos\theta}{a^2\sin^2\theta+b^2\cos^2\theta}=a^2\frac{\sin^3\theta}{\cos\theta}-b^2\frac{\cos^3\theta}{\sin\theta}$$

Derivative vanishes when $$\tan^2\theta=\frac ba$$

Can you finish from here?

Solution 2

As suggested in the comments, I will assume that the question is asking the following: show that the minimal length of a length of a line segment tangent to the ellipse and ending on the two coordinate axes is $a+b$.

Your method is fine. We note that your formula immediately gives the two intercepts: $(0,\frac {b}{sin\theta})$ and $(\frac {a}{cos\theta},0)$. Pythagorus then gives us the length of the associated line segment. Squaring that length we see that we are asked to minimize the function $$f(\theta)=\frac {a^2}{cos^2\theta}+\frac {b^2}{sin^2\theta}$$

A simple derivative calculation shows that this is minimized when $$tan^4\theta=\frac{a^2}{b^2}\;\Rightarrow\;tan^2\theta=\frac{a}{b}$$ Substituting this into the defining equation for $f(\theta)$ we see that $$f(\theta_{min})=a^2(1+\frac ba)+b^2(1+\frac ab)=a^2+2ab+b^2$$ from which the desired result follows at once.

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Author by

Brahmagupta

Updated on August 01, 2022

Comments

• Brahmagupta over 1 year

Prove that the least intercept made on the tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ by the axes is $a+b$.Also find the point of contact of the corresponding tangent.

I tried.Let the point of contact of the tangent with the ellipse be $(a\cos \theta,b\sin\theta)$.Then equation of tangent is $y-b\sin\theta=\frac{-b\cos\theta}{a\sin\theta}(x-a\cos\theta)\Rightarrow bx\cos\theta+ay\sin\theta=ab$

Then how to move ahead and get answer?I stuck,please help.

• ajotatxe about 8 years
I have just realized OP's comment. I leave the answer for other possible readers.