If \( \theta \) is the acute angle between the tangents drawn from the point \( (2, 3) \) to the hyperbola \( 5x^2 - 6y^2 - 30 = 0 \), then \( \tan \theta \) is:
Show Hint
- \( \frac{\pi}{4} \)
- \( \frac{3}{4} \)
- \( \frac{4}{3} \)
- \( \frac{\pi}{2} \)
The Correct Option is C
Approach Solution - 1
To determine \( \tan \theta \), where \( \theta \) is the acute angle between the tangents drawn from the point \( (2, 3) \) to the hyperbola \( 5x^2 - 6y^2 - 30 = 0 \), we follow these steps:
First, rewrite the equation of the hyperbola \( 5x^2 - 6y^2 - 30 = 0 \) in the standard form:
\( \frac{5x^2}{30} - \frac{6y^2}{30} = 1 \) simplifies to \( \frac{x^2}{6} - \frac{y^2}{5} = 1 \).
The general equation of a tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by:
\( \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \).
Substitute \( a^2 = 6 \) and \( b^2 = 5 \) into the tangent equation:
\( \frac{xx_1}{6} - \frac{yy_1}{5} = 1 \).
Let \( (x_1, y_1) = (2, 3) \). The equation of the tangents from the point \( (2, 3) \) is:
\( \frac{2x}{6} - \frac{3y}{5} = 1 \).
This can be rewritten as:
\( \frac{x}{3} - \frac{y}{\frac{5}{3}} = 1 \). Therefore, the slopes of the tangents (denote as \( m_1 \) and \( m_2 \)) can be found using the fact that the product of their slopes is:
\( m_1 \times m_2 = \frac{b^2}{a^2} = \frac{5}{6} \).
For an angle \( \theta \) between the tangents drawn from a point outside the hyperbola, \(\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1m_2} \right| \). Using \( m_1 \times m_2 = \frac{b^2}{a^2} \):
\( \tan \theta = \frac{\sqrt{\left(\frac{a^2 - b^2}{b^2}\right)}}{1 + \frac{b^2}{a^2}} \) is simplified.
Substitute \( a^2 = 6 \) and \( b^2 = 5 \):
\( \tan \theta = \frac{\sqrt{\frac{36 - 25}{25}}}{1 + \frac{5}{6}} = \frac{\sqrt{\frac{11}{25}}}{\frac{11}{6}} = \frac{\sqrt{11} \cdot 6}{11 \cdot 5} = \frac{4}{3} \).
Thus, the value of \(\tan \theta\) is \(\frac{4}{3}\), which is the correct option.
Approach Solution -2
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