A chord PQ of ellipse frac{x^2}{9} + frac{y^2}{4} = 1 subtends right angle at its&nb

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10-2. Parabola, Ellipse, Hyperbola

normal

A chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

a circle

a parabola

an ellipse

a hyperbola

Solution

Let the point of intersection be $\mathrm{R}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) .$ Then $\mathrm{PQ}$ is the chord of contact of the ellipse with respect to $\mathrm{R}$ and its equation will be $\frac{\mathrm{xx}_{1}}{9}+\frac{\mathrm{yy}_{1}}{4}=1$

Now combined equation of lines joining $P, Q$ to centre $\mathrm{O}(0,0)$ is [It is obtained making $\mathrm{x}^{2} / 9+\mathrm{y}^{2} / 4=1$homogeneous with help of $(1)]$

$\frac{x^{2}}{9}+\frac{y^{2}}{4}=\left(\frac{x x_{1}}{9}+\frac{y y_{1}}{4}-1\right)^{2}$

As given $OP$ $\perp$ $OQ,$ so coefficient of

$\mathrm{x}^{2}+$ coefficient of $\mathrm{y}^{2}=0$

$\Rightarrow\left(\frac{x_{1}^{2}}{81}-\frac{1}{9}\right)+\left(\frac{y^{2}}{16}-\frac{1}{4}\right)=0$

Hence $\operatorname{locus}$ of $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)$ is $\frac{\mathrm{x}^{2}}{81}+\frac{\mathrm{y}^{2}}{16}=\frac{13}{36},$ which is an ellipse.

Standard 11

Mathematics

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(JEE MAIN-2023)

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

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(IIT-2023)

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If $C$ is the centre of the ellipse $9x^2 + 16y^2$ = $144$ and $S$ is one focus. The ratio of $CS$ to major axis, is

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A chord $PQ$ of the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$ subtends right angle at its centre. The locus of the point of intersection of tangents drawn at $P$ and $Q$ is-

Solution

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