The asymptote of the hyperbola x2-a2-y2-b2- 1 makes an angle of 60- with x-axis- Write down the equation of diameter conjugate to the diameter y- 2x-

Slope of x/a y/b= 1 is nbsp; b/a= tan 60^∘=√(3) nbsp;Also slope of one diameter is m_1= 2. nbsp;Two diameters are conjugate if Hence the ...

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Standard XII

Mathematics

Definition of Ellipse

The asymptote...

Question

The asymptote of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ makes an angle of $60^{\circ}$ with x-axis. Write down the equation of diameter conjugate to the diameter $y = 2 x .$

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Solution

Slope of $\frac{x}{a} - \frac{y}{b} = 1$ is $\frac{b}{a} = tan 60^{\circ} = \sqrt{3}$ Also slope of one diameter is $m_{1} = 2.$ Two diameters are conjugate if Hence the other diameter is $m_{1} m_{2} = b^{2} / a^{2}$
$∴ 2 m_{2} = 3 ∴ m_{2} = 3 / 2.$
Hence the other diameter is $y = \frac{3}{2} x$ or $2 y - 3 x = 0$

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The combined equation of the asymptotes of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ is $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1$ ..

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The asymptote of the hyperbola x2a2−y2b2=1 makes an angle of 60∘ with x-axis. Write down the equation of diameter conjugate to the diameter y=2x.

Slope of xa−yb=1 is ba=tan60∘=√3 Also slope of one diameter is m1=2. Two diameters are conjugate if Hence the other diameter is m1m2=b2/a2∴2m2=3∴m2=3/2.Hence the other diameter is y=32x or 2y−3x=0

The asymptote of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ makes an angle of $60^{\circ}$ with x-axis. Write down the equation of diameter conjugate to the diameter $y = 2 x .$