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Find the equation of the diameter of hyperbola 16x2 - 9y2 = 144,  which is conjugate to the diameter whose equation is x = 2y.

Option: 1

32y = 9x


Option: 2

32x = 9y


Option: 3

16y = 3x


Option: 4

16x = 3y


Answers (1)

best_answer

 

 

Diameter of Hyperbola and Director Cirlce of Hyperbola -

Diameter of Hyperbola: 

The locus of the mid-points of a system of parallel chords of a hyperbola is called a diameter of the hyperbola.

 

\\\text{The equation of a diameter to a hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\;\text{is }y=\frac{b^{2}}{a^{2} m} x.

Let R (h, k) be the mid-point of the chord y = mx + c of the hyperbola.

 

 

\\\text{Now, using the formula of chord bisected at a given point}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;T=S_1}\\\\\mathrm{\Rightarrow\;\;\;\;\;\;\;\;\;\;\;\;\; \frac{x h}{a^{2}}-\frac{y k}{b^{2}}-1=\frac{h^{2}}{a^{2}}-\frac{k^{2}}{b^{2}}-1}\\\\\mathrm{\therefore \;\;\;\;\;\;\;\;\;\;\;\;\;\;slope=\frac{hb^2}{ka^2}=m}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;k=\frac{b^2h}{a^2m}}\\\\\text{Hence,\;\; the locus of the mid-point is }\;y=\frac{b^{2} x}{a^{2} m}

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Let the equation of the diameter, which is conjugate to x = 2y is y = m1x

As we know that two diameters y = m1x and y = m2x are conjugates, if

\\ {m_{1} m_{2}=\frac{b^{2}}{a^{2}}} \\ {m_{1} \times \frac{1}{2}=\frac{16}{9}} \\ {m_{1}=\frac{32}{9}}

Hence, the equation of the conjugate diameters is y=\frac{32}{9}x

Posted by

Pankaj Sanodiya

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