Maximize the length of common tangent

Geometry Level 4

You are given an ellipse $\dfrac{x^2}{81} + \dfrac{y^2}{16} = 1$ and a family of concentric circles with variable radius $x^2 + y^2 = r^2$ . If the circle having the maximum length of common tangent with ellipse has radius $a$ and the maximum length be $l$ , find $\dfrac{a}{l}$ .

The answer is 1.2.

2 solutions

Indraneel Mukhopadhyaya
Mar 2, 2017

Observe that the problem given to find the maximum length of common tangent is the same as this problem: Let line L be a normal to a standard ellipse at some point on it. Then find the maximum possible distance of origin from L.Take point on ellipse as (acos(t), bsin (t)).Equation of normal at that point is axsec(t)-bycosec(t)=a²-b². Distance of origin from the normal is (a²-b²) divided by square root of [a²(sec(t))^2 + b²(cosec(t))^2]. Now this can be optimised using calculus.We get the maximum value when (tan(t))^2=b/a and the required maximum is (a-b).Using Pythagoras theorem, the radius of circle=distance of the origin from corresponding tangent line is √(ab).

Can you elaborate this statement - " Observe that the problem given to find the maximum length of common tangent is the same as this problem: Let line L be a normal to a standard ellipse at some point on it. "

Kartik Sharma - 4 years, 3 months ago

Suppose that the common tangent intersects ellipse at A and circle at B.Now draw a normal at A to the ellipse.Let foot of perpendicular from origin to the normal at A be point X.We see that quadrilateral ABOX is a rectangle, hence we get AB =OX.But OX is the distance of the origin from the normal at A to the ellipse.Hence,the assertion that I made holds.

Indraneel Mukhopadhyaya - 4 years, 3 months ago

Kartik Sharma
Feb 28, 2017

I will extend this solution later. I am giving the answer right now.

Maximum length = $a - b$

Radius = $\sqrt{ab}$

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

Maximize the length of common tangent

The answer is 1.2.

2 solutions

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