Question

Equation of the tangent to the hyperbola $4x^{2}-9y^{2}=1$ with eccentric angle $\pi / 6$ is

A. $4x+3y=\sqrt{3}$

B. $4x-3y=\sqrt{3}$

C. $3x-4y=\sqrt{3}$

D. $3x-4y=\sqrt{5}$

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Answer

To find the equation of the tangent to the hyperbola $4x^2 - 9y^2 = 1$ with the eccentric angle $\frac{\pi}{6}$, follow these steps:

  1. Convert the given hyperbola into its standard form: $ 4x^2 - 9y^2 = 1 $
    Divide each term by the constant on the right side: $ \frac{4x^2}{1} - \frac{9y^2}{1} = 1 $

  2. Simplify the expression: $ \left(\frac{x}{\frac{1}{2}}\right)^2 - \left(\frac{y}{\frac{1}{3}}\right)^2 = 1 $
    Thus, the equation in standard form is: $ \frac{x^2}{\left(\frac{1}{2}\right)^2} - \frac{y^2}{\left(\frac{1}{3}\right)^2} = 1 $ Here, $a = \frac{1}{2}$ and $b = \frac{1}{3}$.

  3. Use the general equation of the tangent to the hyperbola with a given eccentric angle $\theta$: $$ \frac{x}{a \sec \theta} - \frac{y}{b \tan \theta} = 1 $$

  4. Substitute the values of $a$, $b$, and $\theta = \frac{\pi}{6}$: $$ \frac{x}{\frac{1}{2} \sec \frac{\pi}{6}} - \frac{y}{\frac{1}{3} \tan \frac{\pi}{6}} = 1 $$

  5. Calculate the trigonometric values:

    • $\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}$

    • $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$

  6. Substitute these trigonometric values back into the equation: $$ \frac{x}{\frac{1}{2} \cdot \frac{2}{\sqrt{3}}} - \frac{y}{\frac{1}{3} \cdot \frac{1}{\sqrt{3}}} = 1 $$

  7. Simplify the coefficients:

    • For the $x$ term: $\frac{x}{1/\sqrt{3}} = x \cdot \sqrt{3} = 2x$

    • For the $y$ term: $\frac{y}{1/(3\sqrt{3})} = y \cdot 3\sqrt{3} = \sqrt{3}y$
      $$ 2x - 3y = \sqrt{3} $$

Therefore, the equation of the tangent to the hyperbola at the given eccentric angle $\frac{\pi}{6}$ is: $ 4x - 3y = \sqrt{3} $

Thus, the correct answer is: B. $4x - 3y = \sqrt{3}$


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