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 $a+b$.

Question

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.

Answer 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?

Answer 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.

Answer 3

$$\dfrac{a^2}{\cos^2\theta}+\dfrac{b^2}{\sin^2\theta}$$

$$=a^2(1+\tan^2\theta)+b^2(1+\cot^2\theta)$$

$$=a^2+b^2(a\tan\theta-b\cot\theta)^2+2a\tan\theta\cdot b\cot\theta$$

$$\ge a^2+b^2+2ab$$

The equality occurs when $a\tan\theta-b\cot\theta=0$

Answer 4

For the Minimization of $$\displaystyle f(\theta) =\frac{a^2}{\cos^2 \theta}+\frac{b^2}{\sin^2\theta} = a^2\cdot \sec^2 \theta+b^2\cdot csc^2 \theta$$

so $$f(\theta) = a^2(1+\tan^2 \theta)+b^2(1+\cot^2 \theta) = a^2+b^2+\left[a^2\tan^2 \theta+b^2\cot^2 \theta\right]\geq a^2+b^2+2ab$$

Using $\bf{A.M\geq G.M}$

$$\displaystyle \frac{a^2\tan^2 \theta+b^2\cot^2 \theta}{2}\geq \sqrt{{a^2\tan^2 \theta \cdot b^2 \cot^2 \theta}} = ab$$

So $\displaystyle a^2\tan^2 \theta+b^2\cot^2 \theta \geq 2ab$