If tan θ ₁. tan θ ₂ = -frac{{{a^2}}}{{{b^2}}} then chord joining two points

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10-2. Parabola, Ellipse, Hyperbola

normal

If $ \tan\ \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$ then the chord joining two points $\theta _1 \& \theta _2$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$ will subtend a right angle at :

focus

centre

end of the major axis

end of the minor axis

Solution

$m1 = \frac{b}{a} \tan \theta _1 ; m_2 = \frac{b}{a} \tan \theta _2$
$\Rightarrow$ $ m_1m_2 =\frac{{{b^2}}}{{{a^2}}} \tan \theta _1 \tan \theta _2 = – 1$
where $m_1$ = slope of $ O$

Standard 11

Mathematics

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(KVPY-2018)

If $ \tan\ \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$ then the chord joining two points $\theta _1 \& \theta _2$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$ will subtend a right angle at :

Solution

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