Side Lengths of Right Angled Triangle from Hyperbola Tangent

To determine which set of values CANNOT form the side lengths of a right-angled triangle, we first need to find the value of $a$ using the condition of tangency for the hyperbola.

Step 1: Find the value of $a$

The given equation of the hyperbola is: $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$ Here, $b^2 = 16$.

The given line is: $2x - y + 1 = 0 \implies y = 2x + 1$ Comparing this with the slope-intercept form $y = mx + c$, we get: $m = 2 \quad \text{and} \quad c = 1$

The condition for the line $y = mx + c$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is: $c^2 = a^2 m^2 - b^2$

Substituting the known values: $1^2 = a^2(2)^2 - 16$ $1 = 4a^2 - 16$ $4a^2 = 17 \implies a^2 = \frac{17}{4}$

Taking the positive square root for side length parameter: $a = \frac{\sqrt{17}}{2} \implies 2a = \sqrt{17}$

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Step 2: Test each option for the right-angled triangle condition

For three side lengths to form a right-angled triangle, the square of the longest side must equal the sum of the squares of the other two sides (Pythagorean theorem).

• Option (A): $a, 4, 1$

  • Side lengths are $\frac{\sqrt{17}}{2}, 4, 1$.
  • Squares of the side lengths are: $\left(\frac{\sqrt{17}}{2}\right)^2 = \frac{17}{4} = 4.25, \quad 4^2 = 16, \quad 1^2 = 1$
  • Checking Pythagoras' theorem: $4.25 + 1 = 5.25 \neq 16$
  • Result: These side lengths CANNOT form a right-angled triangle.

• Option (B): $a, 4, 2$

  • Side lengths are $\frac{\sqrt{17}}{2}, 4, 2$.
  • Squares of the side lengths are: $\left(\frac{\sqrt{17}}{2}\right)^2 = \frac{17}{4} = 4.25, \quad 4^2 = 16, \quad 2^2 = 4$
  • Checking Pythagoras' theorem: $4.25 + 4 = 8.25 \neq 16$
  • Result: These side lengths CANNOT form a right-angled triangle.

• Option (C): $2a, 8, 1$

  • Side lengths are $\sqrt{17}, 8, 1$.
  • Squares of the side lengths are: $(\sqrt{17})^2 = 17, \quad 8^2 = 64, \quad 1^2 = 1$
  • Checking Pythagoras' theorem: $17 + 1 = 18 \neq 64$
  • Result: These side lengths CANNOT form a right-angled triangle.

• Option (D): $2a, 4, 1$

  • Side lengths are $\sqrt{17}, 4, 1$.
  • Squares of the side lengths are: $(\sqrt{17})^2 = 17, \quad 4^2 = 16, \quad 1^2 = 1$
  • Checking Pythagoras' theorem: $4^2 + 1^2 = 16 + 1 = 17 = (2a)^2$
  • Result: These side lengths CAN form a right-angled triangle.

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Conclusion

The options that CANNOT form the side lengths of a right-angled triangle are (A), (B), and (C).