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  • Find the locus of point of intersection of pair of tangents to the ellipse if the sum of the ordinates of the point of contact is b.Option: 1 <img al

Find the locus of point of intersection of pair of tangents to the ellipse if the sum of the ordinates of the point of contact is b.

Option: 1

\left ( \frac{x^2 }{a^2} + \frac{y^2}{b^2 } \right ) b / 4y = 1


Option: 2

\left ( \frac{x^2}{a^2}+\frac{y^2}{b^2} \right )b/2y= 1


Option: 3

\left ( \frac{x^2}{a^2}+\frac{y^2}{b^2} \right )2b/y= 1


Option: 4

\left ( \frac{x^2}{a^2}+\frac{y^2}{b^2} \right )b/2y=4


Answers (1)

best_answer

 

 

Pair of Tangent -

Pair of Tangent:
 

\\ {\text {The equation of pair of tangent from the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { to the Ellipse }} \\ {\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \text { is }\left(\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}-1\right)\left(\frac{\mathrm{x}_{1}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}_{1}^{2}}{\mathrm{b}^{2}}-1\right)=\left(\frac{\mathrm{x} \mathrm{x}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{y} \mathrm{y}_{1}}{\mathrm{b}^{2}}-1\right)^{2}} \\ {\text { or, } \mathrm{SS}_{1}=\mathrm{T}^{2}} \\ {\text { where, } \mathrm{S}=\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}-1} \\\\ {\mathrm{S}_{1}=\frac{\mathrm{x}_{1}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}_{1}^{2}}{\mathrm{b}^{2}}-1} \\\\ {\mathrm{T}=\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{b}^{2}}-1}

Where points Q and R are the point of contacts of the tangents to the ellipse.

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- wherein

 

 

b \sin \alpha + b \sin \beta = b \\\\ \sin \alpha + \sin \beta = 1 \\\\ h = \frac{a \cos \frac{\alpha +\beta }{2}}{\cos \frac {\alpha -\beta }{2}} .... (1 ) \\\\ k = \frac{b \sin \frac{\alpha +\beta }{2}}{\cos \frac {\alpha -\beta }{2}} .... (2)\\\\ (1)^2 + 2^2 \\\\ \frac{h^2}{a^2 }+ \frac{k^2}{b^2 } = \frac{1}{\cos ^2 \frac{\alpha -\beta }{2}}

\rightarrow \left ( \frac{h^2}{k^2} + \frac{k^2}{b^2}\right ) \left ( \cos^2 \frac{\alpha -\beta }{2} \right ) = 1 \\\\ 2 \sin \frac{\alpha +\beta }{2} \cos \frac{\alpha -\beta }{2} = 1 \\\\ from (2) and (4) \\\\ \frac{k \cos \left ( \frac{\alpha -\beta }{2} \right )}{b} = \frac{1}{2 \cos\left ( \frac{\alpha -\beta}{2} \right )}

\frac{2k}{b} \cos ^2 \left ( \frac{\alpha -\beta }{2} \right ) = 1 .... (5) \\\\ from (3) and (5 )

\left ( \frac{x^2}{a^2}+\frac{y^2}{b^2} \right )b/2y= 1

Posted by

manish painkra

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