Question

If the line \(y = 2x+λ\) is a tangent to the hyperbola \(36x^2-25y^2 = 3600\), then \(λ =\)

• A. \(\pm 36\)
• B. \(\pm 25\)
• C. \(\pm 16\)
• D. \(\pm 9\)

Hint:

Reduce the hyperbola to standard form and use c^2 = a^2 m^2 - b^2.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
Divide \(36x^2 - 25y^2 = 3600\) by 3600 to get \(\dfrac{x^2}{100} - \dfrac{y^2}{144} = 1\). So \(a^2 = 100\) and \(b^2 = 144\).

Step 2: Tangency condition:
The line \(y = mx + c\) touches \(\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1\) when \(c^2 = a^2m^2 - b^2\).

Step 3: Apply:
Here \(m = 2\), so
\[ c^2 = 100(4) - 144 = 256 \Rightarrow c = \pm 16 \]
So \(\lambda = \pm 16\). The values 36, 25 and 9 do not satisfy \(c^2 = 256\).

Step 4: Check:
Substituting \(y = 2x + 16\): \(36x^2 - 25(4x^2 + 64x + 256) = 3600\) gives \(-64x^2 - 1600x - 10000 = 0\), i.e. \(x^2 + 25x + 156.25 = 0\), with discriminant \(625 - 625 = 0\). This confirms tangency.

Final Answer:
The tangent condition gives lambda = plus or minus 16. \[ \boxed{\text{(C) }\pm16} \]