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Tangents PA and PB are drawn from a point P to the ellipse. Find the locus of a point P, if the area of the triangle formed by the chord of contact AB and axes of co-ordinates are constant (c).

Option: 1

x\cdot y=\frac{4a^{2} b^{2}}{c}


Option: 2

x\cdot y=\frac{2a^{2} b^{2}}{c}


Option: 3

x\cdot y=\frac{a^{2} b^{2}}{2c}


Option: 4

x\cdot y=\frac{a^{2} b^{2}}{c}


Answers (1)

best_answer

 

 

Chord of Contact -

Chord of Contact:

\\ {\text {The equation of chord of contact of tangents from the point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \text { to }} \\ {\text { the Ellipse } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \text { is } \frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{b}^{2}}=1}.

Equation of Chord bisected at a given point:

\\ {\text {The equation of chord of the ellipse } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1 \text { bisected at a given point } \mathrm{P}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)} \\ {\text { is } \frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}+\frac{\mathrm{yy}_{1}}{\mathrm{b}^{2}}-1=\frac{\mathrm{x}_{1}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y_1}^{2}}{\mathrm{b}^{2}}-1} \\\\ {\text { or, } \mathrm{T}=\mathrm{S}_{1}}

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Let the coordinates of P be (h, k)

The equation of its chord of contact with respect to the ellipse \frac{x^2}{a^2}+\frac{ y^2}{b^2}=1 is

\frac{hx}{a^2}+\frac{ ky}{b^2}=1

It meets the axes in A\left(\frac{a^2}{h},0 \right ) and B \left(0,\frac{b^2}{k} \right )

Now, area of the triangle OAB is 

=\frac{1}{2} \cdot O A \cdot O B

\\ {=\frac{1}{2} \cdot \frac{a^{2}}{h} \cdot \frac{b^{2}}{k}}

\\ {=\frac{a^{2} b^{2}}{2 \cdot h \cdot k}}

=\text { constant }=c

Hence, the locus of (h, k) is x\cdot y=\frac{a^{2} b^{2}}{2c}

Posted by

Ritika Jonwal

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