Question

The foci of a hyperbola are (±2,0) and its eccentricity is \(\frac{3}{2}\) . A tangent, perpendicular to the line 2x + 3y = 6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a| + |5b| is equal to_____.

Answer 1

Correct Answer: 12

We are given a hyperbola with foci \( (\pm 2, 0) \) and its eccentricity \( e = \frac{3}{2} \). The standard form of a hyperbola is: \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \] where \( c \) is the focal distance, and the relation between \( a \), \( b \), and \( c \) is: \[ c^2 = a^2 + b^2 \] From the problem, the foci are \( (\pm 2, 0) \), so we know: \[ c = 2 \] The eccentricity is given as \( e = \frac{3}{2} \). From the formula for eccentricity: \[ e = \frac{c}{a} \Rightarrow \frac{3}{2} = \frac{2}{a} \] Solving for \( a \): \[ a = \frac{4}{3} \] Next, we use the relation \( e^2 = 1 + \frac{b^2}{a^2} \) to find \( b \): \[ e^2 = \left( \frac{3}{2} \right)^2 = \frac{9}{4} \] \[ \frac{9}{4} = 1 + \frac{b^2}{a^2} \Rightarrow \frac{9}{4} - 1 = \frac{b^2}{a^2} \] \[ \frac{5}{4} = \frac{b^2}{a^2} \Rightarrow b^2 = \frac{5}{4} \times \left( \frac{4}{3} \right)^2 = \frac{5}{4} \times \frac{16}{9} = \frac{20}{9} \] Thus, we find: \[ b = \frac{\sqrt{20}}{3} = \frac{2\sqrt{5}}{3} \] Now, a tangent to the hyperbola is drawn at a point in the first quadrant. We are given that the tangent is perpendicular to the line \( 2x + 3y = 6 \). The slope of this line is: \[ m_{\text{line}} = -\frac{2}{3} \] Since the tangent is perpendicular to this line, the slope of the tangent line, \( m_{\text{tangent}} \), will be the negative reciprocal of the slope of the given line: \[ m_{\text{tangent}} = \frac{3}{2} \] The equation of the tangent to the hyperbola at the point \( (x_0, y_0) \) is: \[ y - y_0 = m_{\text{tangent}} (x - x_0) \] Substituting \( m_{\text{tangent}} = \frac{3}{2} \), the equation becomes: \[ y - y_0 = \frac{3}{2} (x - x_0) \] The intercepts on the x-axis (\( x \)) and y-axis (\( y \)) can be found by setting \( y = 0 \) and \( x = 0 \), respectively. By solving the equation of the tangent for \( x \) and \( y \), we obtain the intercepts: \[ x = \frac{8}{9}, \quad y = \frac{4}{3} \] Thus, the intercepts on the x and y axes are \( a = \frac{8}{9} \) and \( b = \frac{4}{3} \), respectively. Finally, we calculate: \[ |6a| + |5b| = 6 \times \frac{8}{9} + 5 \times \frac{4}{3} = \frac{48}{9} + \frac{20}{3} = \frac{48}{9} + \frac{60}{9} = \frac{108}{9} = 12 \] Thus, the value of \( |6a| + |5b| \) is \( \boxed{12} \).