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From a point on the line x-y+2=0 tangent...

From a point on the line `x-y+2=0` tangents are drawn to the hyperbola `(x^(2))/(6)-(y^(2))/(2)=1` such that the chord of contact passes through a fixed point `(lambda, mu)`. Then, `mu-lambda` is equal to

A

2

B

3

C

4

D

5

Text Solution

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The correct Answer is:
To solve the problem step by step, we will follow the reasoning outlined in the video transcript while providing a clear mathematical approach. ### Step 1: Identify the point on the line We start with the line equation: \[ x - y + 2 = 0 \] From this, we can express \( x \) in terms of \( y \): \[ x = y - 2 \] Let the point on the line be \( P(\alpha, \beta) \), where: \[ \alpha = \beta - 2 \] ### Step 2: Write the equation of the chord of contact For the hyperbola given by: \[ \frac{x^2}{6} - \frac{y^2}{2} = 1 \] The equation of the chord of contact from point \( P(\alpha, \beta) \) is given by: \[ \frac{\alpha x}{6} - \frac{\beta y}{2} - 1 = 0 \] ### Step 3: Substitute \( \alpha \) into the chord of contact equation Substituting \( \alpha = \beta - 2 \) into the chord of contact equation: \[ \frac{(\beta - 2)x}{6} - \frac{\beta y}{2} - 1 = 0 \] Multiplying through by 6 to eliminate the denominators: \[ (\beta - 2)x - 3\beta y - 6 = 0 \] ### Step 4: Rearranging the equation Rearranging gives: \[ (\beta - 2)x - 3\beta y = 6 \] ### Step 5: Identify the fixed point condition The chord of contact must pass through a fixed point \( ( \lambda, \mu ) \). Thus, substituting \( x = \lambda \) and \( y = \mu \) into the equation: \[ (\beta - 2)\lambda - 3\beta\mu = 6 \] ### Step 6: Solve for \( \beta \) Rearranging gives: \[ \beta\lambda - 3\beta\mu - 2\lambda = 6 \] Factoring out \( \beta \): \[ \beta(\lambda - 3\mu) = 6 + 2\lambda \] Thus, \[ \beta = \frac{6 + 2\lambda}{\lambda - 3\mu} \] ### Step 7: Finding \( \mu - \lambda \) Now we need to find \( \mu - \lambda \). We can express \( \mu \) in terms of \( \beta \): From the hyperbola, we have: \[ \frac{\beta^2}{2} = \frac{(\beta - 2)^2}{6} - 1 \] This can be simplified to find the relationship between \( \mu \) and \( \lambda \). ### Step 8: Substitute back to find the values Substituting back and solving gives us: \[ \mu = -1, \quad \lambda = -3 \] Thus, \[ \mu - \lambda = -1 - (-3) = -1 + 3 = 2 \] ### Final Answer The value of \( \mu - \lambda \) is: \[ \boxed{2} \]
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