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  • If     be two points on  <img alt="\mathrm {b^2 x^2+a^2 y^2=a^2 b^2 }" src

If   \alpha, \beta  be two points on  \mathrm {b^2 x^2+a^2 y^2=a^2 b^2 }  such that  \mathrm {b^2 \tan \alpha+a^2 \cot \beta=0, } then chord joining   \mathrm {\alpha \: \: \& \: \: \beta } will subtend  \mathrm {90^{\circ} } at
 

Option: 1

focus
 


Option: 2

directrix
 


Option: 3

vertex
 


Option: 4

centre


Answers (1)

best_answer

\mathrm {b^2 x^2+a^2 y^2=a^2 b^2 \Rightarrow \frac{x^2}{a^2}+\frac{y^2}{b^2}=1} 
At  \alpha, \beta  means at  \mathrm {P(a \cos \alpha, b \sin \alpha)\: \: \& \: \: Q(a \cos \beta, b \sin \beta)}
Let C be the centre i.e. (0,0)
\therefore  Slope of C P \times  slope of C Q

\begin{aligned} & =\mathrm{\frac{b \sin \alpha-0}{a \cos \alpha-0} \times \frac{b \sin \beta-0}{a \cos \beta-0}} \\\\ & =\mathrm {\frac{b^2}{a^2} \tan \alpha \cdot \tan \beta=\frac{b^2 \tan \alpha}{a^2 \cot \beta}=\frac{-a^2 \cot \beta}{a^2 \cot \beta}=-1} \end{aligned}

\therefore \quad PQ subtends 90o at centre.

Posted by

Divya Prakash Singh

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