Let the values of p , for which the shortest distance between the lines (x+1)/3=y/4=z/5 and r⃗=(p î+2 ĵ+k̂)+λ(2 î+3 ĵ+4 k̂) is 1/√6, be a, b,(a<b). Then the length of the latus rectum of the ellipse x²/a²+y²/b²=1 is :

Solution

Maths sir · 4 steps

Find p from the shortest distance formula, then check which axis of the ellipse is the major one.

1. Read the lines. Line 1 passes through A(-1, 0, 0) with direction d₁⃗ = (3, 4, 5). Line 2 passes through B(p, 2, 1) with direction d₂⃗ = (2, 3, 4).
2. Common perpendicular. d₁⃗ × d₂⃗ = (4 · 4 - 5 · 3, 5 · 2 - 3 · 4, 3 · 3 - 4 · 2) = (1, -2, 1), with magnitude √6. Also AB⃗ = (p+1, 2, 1), so AB⃗ · (1, -2, 1) = p + 1 - 4 + 1 = p - 2.

   (|p-2|)/√6 = 1/√6 ⇒ |p-2| = 1 ⇒ p = 1 or 3

3. The ellipse. With a = 1 and b = 3, the ellipse is x²/1 + y²/9 = 1. Since 9 > 1, the major axis lies along the y-axis. The semi-major axis is 3 and the semi-minor axis is 1.
4. So. Latus rectum = (2 × (semi-minor)²)/(semi-major) = (2 × 1²)/3 = 2/3, which is option (d).

The idea to keep. For skew lines, project AB⃗ onto d₁⃗ × d₂⃗. For an ellipse, latus rectum is 2 × (semi-minor)² ÷ semi-major, so first see which axis is longer.

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