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: For a hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\), line \(y=mx+c\) is tangent if \(c^2=a^2m^2-b^2\). Use the point \(P(x_1,-1)\) to express \(c\), form a quadratic in \(m\), and use sum of roots.

Step 1: Tangent through \(P(x_1,-1)\): \(y+1=m(x-x_1) \Rightarrow y=mx - (mx_1+1)\). So \(c=-(mx_1+1)\). Hyperbola: \(a^2=2, b^2=3\).

Step 2: Tangency condition: \((mx_1+1)^2 = 2m^2 - 3 \Rightarrow (x_1^2-2)m^2 + 2x_1m + 4 = 0\).

Step 3: Sum of slopes = \(-\frac{2x_1}{x_1^2-2} = 2 \Rightarrow -2x_1 = 2x_1^2-4 \Rightarrow x_1^2+x_1-2=0 \Rightarrow (x_1+2)(x_1-1)=0\).

Step 4: \(x_1<0 \Rightarrow x_1=-2\).

Step 5: Write the final answer. \(\boxed{-2}\)