Another Ellipse Problem

Geometry Level 4

A variable point $P$ on an ellipse of eccentricity $e=\dfrac{1}{8}$ , is joined to it's focii $S_1$ and $S_2$ .

Given that the locus of the incentre of the triangle $\Delta P S_1 S_2$ comes out to be a conic;

Evaluate its eccentricity $e'$ . Now $e'$ is of the form $\frac{\sqrt{a}}{b}$ , where $a$ and $b$ are coprime positive integers, find $a\times b$ .

The answer is 6.

1 solution

Ujjwal Rane
Feb 12, 2015

My apologies for formatting. I was not getting the scroll bar in this window and thus getting back from Preview to Edit was painful. So typed without much feedback :-(

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Let the original ellipse with eccentricity $\frac{1}{8}$ be $\frac{x^2}{64}+\frac{y^2}{63}$ giving a = 8, b = $\sqrt{63}$ and f = 1

For any point P(h,k), the described in-center C will be on the angle bisector of $\angle S_{1}PS_{2}$ . But such angle bisector is also the normal at P!

Equation of a normal at P(h,k) will be $y = \frac{64k}{63h}x-\frac{k}{63}$

Radius of incircle $R = \frac{areaOfS_{1}PS_{2}}{semiperimeter OfS_{1}PS_{2}} = \frac{fk}{f+a}=\frac{k}{9}$ = y coordinate of the incenter

Substituting this in equation for the normal: gives $C=(\frac{h}{8},\frac{k}{9})$

Thus the locus of center C is an ellipse: $\frac{x^2}{1}+\frac{81y^2}{63}$ giving a = 1 and $b = \frac{\sqrt{63}}{9}$ and eccentricity $e' = \sqrt{1-\frac{63}{81}}=\frac{\sqrt{2}}{3}$

Thus the quantity asked is $2 \times 3 = 6$

Nice work $\ddot\smile$

A Former Brilliant Member - 6 years, 4 months ago

How do we prove the angle bisector is normal at the point ??

Vishal Yadav - 4 years, 2 months ago

That is the property of ellipse. The normal is the bisector. Since an angle bisector is a unique line, the property can be applied in the other direction too. i.e. bisector is the normal.

Ujjwal Rane - 4 years, 1 month ago

Another Ellipse Problem

The answer is 6.

1 solution

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