The locus of the point of intersection of the lines ( √3 )kx + ky - 4√3 = 0 and √3 x - y - 4( √3 )k = 0 is a conic, whose eccentricity is _________.

Solution

Maths sir · 4 steps

Do not solve for x and y separately: multiply the two equations so that k cancels out.

1. Rewrite both lines. The first line is k(√3x + y) = 4√3 and the second is √3x - y = 4√3k. The point of intersection satisfies both.
2. Eliminate k. Multiply the two equations side by side. The left gives k(√3x + y)(√3x - y) and the right gives 4√3 × 4√3k = 48k. Divide by k, which is not zero.

   3x² - y² = 48 ⇒ x²/16 - y²/48 = 1

3. Find eccentricity. This is a hyperbola with a² = 16 and b² = 48. So e² = 1 + b²/a² = 1 + 48/16 = 4.
4. So. The eccentricity is e = √4 = 2, so the answer is 2.

The common slip: 4. Stopping at e² = 1 + 48/16 and reporting that value gives 4. Take the square root to get e.

The idea to keep. For a locus with a parameter, combine the equations so that the parameter disappears. Here one line has k and the other has 1/k in effect, so multiplying them kills it.

Answer from the NTA final answer key. Solution by PrepWiser.