Point 'O' is centre of ellipse with major axis AB & minor axis CD . Point F is one focus

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10-2. Parabola, Ellipse, Hyperbola

normal

Point $'O' $ is the centre of the ellipse with major axis $AB$ $ \&$ minor axis $CD$. Point $F$ is one focus of the ellipse. If $OF = 6 $ $ \&$ the diameter of the inscribed circle of triangle $OCF$ is $2, $ then the product $ (AB)\,(CD) $ is equal to

$65$

$52$

$78$

none

Solution

$a^2 e^2 = 36 \Rightarrow a^2 – b^2 = 36….. (1)$
Using $r = (s – a) \tan \frac {A}{2}\, in\, \Delta OCF $
$1 = (s – a) tan 45°$ when $a = CF$
$2 = 2 (s – a)$
$= 2s – 2a = 2s – AB $
$= (OF + FC + CO) – AB $
$2 = 6 +\frac{{A\,B}}{2} + \frac{{C\,D}}{2} – AB$
$ \frac{AB-CD}{2} = 4 \Rightarrow 2 (a – b) = 8 \Rightarrow a – b = 4…. (2)$
From $(1)$ and $ (2)$ $ a + b = 9$

$ \Rightarrow $ $2a = 13$ $;$ $ 2b = 5$

$\Rightarrow $ $(AB) (CD) = 65$

Standard 11

Mathematics

The eccentricity of an ellipse, with its centre at the origin, is $\frac{1}{2}$. If one of the directrices is $x = 4$, then the equation of the ellipse is

easy

(AIEEE-2004)

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Another ellipse $E _2$ passing through the point $(0,4)$ circumscribes the rectangle $R$.. The eccentricity of the ellipse $E _2$ is

normal

(IIT-2012)

What will be the equation of that chord of ellipse $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{9} = 1$ which passes from the point $(2,1)$ and bisected on the point

hard

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$

medium

Let $f(x)=x^2+9, g(x)=\frac{x}{x-9}$ and $\mathrm{a}=\mathrm{fog}(10), \mathrm{b}=\operatorname{gof}(3)$. If $\mathrm{e}$ and $1$ denote the eccentricity and the length of the latus rectum of the ellipse $\frac{x^2}{a}+\frac{y^2}{b}=1$, then $8 e^2+1^2$ is equal to.

hard

(JEE MAIN-2024)