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: Substitute and use the discriminant:
Put $y = 2x + \lambda$ into $36x^2 - 25y^2 = 3600$: $36x^2 - 25(4x^2 + 4\lambda x + \lambda^2) - 3600 = 0$.

Step 2: Simplify:
$-64x^2 - 100\lambda x - 25\lambda^2 - 3600 = 0$, or $64x^2 + 100\lambda x + 25\lambda^2 + 3600 = 0$.

Step 3: Discriminant zero:
$10000\lambda^2 - 4\cdot64(25\lambda^2 + 3600) = 0 \Rightarrow 3600\lambda^2 = 921600 \Rightarrow \lambda^2 = 256$. So $\lambda = \pm16$, option (C).

Final Answer:
Lambda equals plus or minus 16. \[ \boxed{\text{(C) }\pm16} \]