Question

From a point \[ P(x_1,-1), \qquad (x_1<0), \] two tangents are drawn to the hyperbola \[ \frac{x^2}{2}-\frac{y^2}{3}=1. \] If the sum of the slopes of the tangents is \(2\), then \(x_1=\)

• A. \(-1\)
• B. \(-2\)
• C. \(-\dfrac13\)
• D. \(-\dfrac25\) \bigskip

Hint:

For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] a line \(y=mx+c\) is tangent iff \[ c^2=a^2m^2-b^2. \] When tangents are drawn from a point, substitute the point into the line equation and obtain a quadratic in \(m\). The roots give the slopes of the tangents.

Answer 1

The Correct Option is B

Concept: A line with slope \(m\) passing through a point \((x_1,y_1)\) is \[ y-y_1=m(x-x_1). \] For the hyperbola \[ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, \] the line \[ y=mx+c \] is a tangent if \[ c^2=a^2m^2-b^2. \]

Step 1: Write the equation of a tangent through \(P(x_1,-1)\). A line through \(P(x_1,-1)\) having slope \(m\) is \[ y+1=m(x-x_1). \] \[ y=mx-(mx_1+1). \] Thus, \[ c=-(mx_1+1). \]

Step 2: Apply the tangency condition. Given hyperbola \[ \frac{x^2}{2}-\frac{y^2}{3}=1, \] so \[ a^2=2, \qquad b^2=3. \] Using \[ c^2=a^2m^2-b^2, \] \[ (mx_1+1)^2=2m^2-3. \] Expanding, \[ m^2x_1^2+2x_1m+1=2m^2-3. \] \[ (x_1^2-2)m^2+2x_1m+4=0. \] \[ \cdots (1) \] The two roots of (1) are the slopes of the two tangents.

Step 3: Use the given sum of slopes. For the quadratic \[ (x_1^2-2)m^2+2x_1m+4=0, \] sum of roots is \[ -\frac{2x_1}{x_1^2-2}. \] Given that the sum of the slopes is \[ 2, \] therefore \[ -\frac{2x_1}{x_1^2-2}=2. \] \[ -x_1=x_1^2-2. \] \[ x_1^2+x_1-2=0. \] \[ (x_1+2)(x_1-1)=0. \] \[ x_1=-2 \quad \text{or} \quad x_1=1. \]

Step 4: Use the condition \(x_1<0\). Since \[ x_1<0, \] the admissible value is \[ x_1=-2. \]

Step 5: Write the final answer. \[ \boxed{-2} \]