Question

The slope of the tangent drawn from the point \( (1,1) \) to the hyperbola \( 2x^2 - y^2 = 4 \) is:

• A. 2
• B. \( \frac{-2 \pm \sqrt{6}}{2} \)
• C. \( -1 \pm \sqrt{6} \)
• D. \( \frac{-2 \pm \sqrt{3}}{2} \) \bigskip

Hint:

When solving tangent problems for hyperbolas, remember to use the parametric form of the hyperbola equation to relate \( x_1 \) and \( y_1 \), and the relationship between the slope and the point on the curve.

Answer 1

The Correct Option is C

We are given the hyperbola equation: \[ 2x^2 - y^2 = 4 \] and the point \( (1,1) \). Step 1: Equation of the tangent to the hyperbola For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equation of the tangent at the point \( (x_1, y_1) \) is: \[ \frac{x_1 x}{a^2} - \frac{y_1 y}{b^2} = 1 \] For the given hyperbola \( 2x^2 - y^2 = 4 \), we can write it in the standard form as: \[ \frac{x^2}{2} - \frac{y^2}{4} = 1 \] Thus, \( a^2 = 2 \) and \( b^2 = 4 \). The equation of the tangent to the hyperbola at any point \( (x_1, y_1) \) will be: \[ \frac{x_1 x}{2} - \frac{y_1 y}{4} = 1 \] Step 2: Slope of the tangent The slope \( m \) of the tangent line at any point \( (x_1, y_1) \) is given by \( -\frac{a^2 y_1}{b^2 x_1} \). Using the values \( a^2 = 2 \) and \( b^2 = 4 \), we have: \[ m = -\frac{2y_1}{4x_1} = -\frac{y_1}{2x_1} \] Step 3: Using the point \( (1,1) \) Now, we know that the tangent line passes through \( (1,1) \). So, the equation of the tangent line at \( (1,1) \) is: \[ \frac{x_1 \cdot 1}{2} - \frac{y_1 \cdot 1}{4} = 1 \] Simplifying: \[ \frac{x_1}{2} - \frac{y_1}{4} = 1 \] Multiplying through by 4 to eliminate fractions: \[ 2x_1 - y_1 = 4 \] Thus, the relation between \( x_1 \) and \( y_1 \) is \( 2x_1 - y_1 = 4 \). Step 4: Solving for the values of \( x_1 \) and \( y_1 \) Now, substituting this relation into the equation of the hyperbola \( 2x_1^2 - y_1^2 = 4 \), we substitute \( y_1 = 2x_1 - 4 \) into this equation. \[ 2x_1^2 - (2x_1 - 4)^2 = 4 \] Expanding the terms: \[ 2x_1^2 - (4x_1^2 - 16x_1 + 16) = 4 \] Simplifying: \[ 2x_1^2 - 4x_1^2 + 16x_1 - 16 = 4 \] \[ -2x_1^2 + 16x_1 - 16 = 4 \] Rearranging: \[ -2x_1^2 + 16x_1 - 20 = 0 \] Dividing through by -2: \[ x_1^2 - 8x_1 + 10 = 0 \] Solving this quadratic equation using the quadratic formula: \[ x_1 = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(10)}}{2(1)} \] \[ x_1 = \frac{8 \pm \sqrt{64 - 40}}{2} = \frac{8 \pm \sqrt{24}}{2} = \frac{8 \pm 2\sqrt{6}}{2} \] \[ x_1 = 4 \pm \sqrt{6} \] Substitute \( x_1 = 4 \pm \sqrt{6} \) into \( y_1 = 2x_1 - 4 \): \[ y_1 = 2(4 \pm \sqrt{6}) - 4 = 8 \pm 2\sqrt{6} - 4 = 4 \pm 2\sqrt{6} \] Step 5: Calculating the slope Finally, the slope \( m \) of the tangent at \( (1,1) \) is: \[ m = -\frac{y_1}{2x_1} = -\frac{4 \pm 2\sqrt{6}}{2(4 \pm \sqrt{6})} \] After simplification: \[ m = -1 \pm \sqrt{6} \] Thus, the correct value for the slope is \( -1 \pm \sqrt{6} \). \bigskip