Cool Ellipse

Geometry Level 2

Point $O$ is the center of the ellipse with major axis $AB$ & minor axis $CD$ Point F is one of the focus of this ellipse.

If $OF=6$ , and the diameter of inscribed circle of triangle $\triangle{OCF}$ is $2$ , then find $(AB)\cdot (CD)$

The answer is 65.

2 solutions

Kushal Patankar
Jan 18, 2015

IMG

Inradius of $(\triangle OCF) = \frac{6b}{\sqrt{36 + b^2} + 6 + b} = 1$

Here we get $b=2.5$

We know by the properties of standard ellipse that focus has coordinates (ae,0) & (-ae,0) .

Given in question that $OF = 6= ae$

And also $b^2 = a^2(1-e^2)$

Solving these give $a=6.5$

Now, $AB = 2a & CD = 2b$

$AB\cdot CD = 13\cdot 5=\boxed{65}$

hey man how you doing

Aashay Godbole - 6 months, 1 week ago

Laurent Shorts
Feb 18, 2017

Inscribed circle has center at $(1;-1$ ) and radius is 1.

Tangents from $F$ to the circle with slope $m$ are $y+1=m(x-1)\pm1·\sqrt{1+m^2}$ . Going through $F=(6;0)$ : $0+1=m(6-1)\pm1·\sqrt{1+m^2}~\implies~m(m-\frac{5}{12})=0$

From the slope $\frac{5}{12}$ , we get $OC=\frac{5}{2}\implies CD=5$ . By Pythagoras, $CF=\frac{13}{2}$ . In an ellipse, $AB=2·CF$ , so $AB=13$ and the answer is $5·13=\boxed{65}$ .

Why +-1.√(1+m²)

kasi viswanath boddeti - 2 years, 2 months ago

This is a formula to find the tangents from $P$ to circle centered in $C$ and with radius $r$ :

$P_y-C_y = m(P_x-C_x)\pm r·\sqrt{1+m^2}$

You can get it from the distance formula between the center $C$ and a line $t$ passing through $P$ . $t: y-P_y=m(x-P_x) \Longleftrightarrow m·x-y+(P_y-m·P_x)=0$ $r = \delta(C, t) = \dfrac{|m·C_x-C_y+(P_y-m·P_x)|}{\sqrt{m^2+(-1)^2}}$ $r = \pm\dfrac{m·C_x-C_y+P_y-m·P_x}{\sqrt{1+m^2}}$ $\pm r·\sqrt{1+m^2} = m·C_x-C_y+P_y-m·P_x$ $m(P_x-C_x)\pm r·\sqrt{1+m^2}=P_y-C_y$

Laurent Shorts - 2 years, 1 month ago

Cool Ellipse

The answer is 65.

2 solutions

1 pending report

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