JEE Main 2022 · 28 July · morning · Single correct · +4 / −1
The foot of the perpendicular from a point on the circle x²+y²=1, z=0 to the plane 2 x+3 y+z=6 lies on which one of the following curves?
- (a) (6 x+5 y-12)²+4(3 x+7 y-8)²=1, z=6-2 x-3 y
- (b) (5 x+6 y-12)²+4(3 x+5 y-9)²=1, z=6-2 x-3 y
- (c) (6 x+5 y-14)²+9(3 x+5 y-7)²=1, z=6-2 x-3 y
- (d) (5 x+6 y-14)²+9(3 x+7 y-8)²=1, z=6-2 x-3 y
Answer and solution
Answer (b) (5 x+6 y-12)²+4(3 x+5 y-9)²=1, z=6-2 x-3 y
Solution
Maths sir · 4 steps
Write the foot in terms of the circle point, then invert to get the circle point back.
- Set up. Take a point P(α, β, 0) on the circle, so α² + β² = 1. The normal to the plane is (2, 3, 1), so the perpendicular from P runs along it. The foot Q(x, y, z) is P moved by t times the normal.
x = α + 2t, y = β + 3t, z = t
- Invert. Since z = t, we get α = x - 2z and β = y - 3z. The foot Q lies on the plane, so z = 6 - 2x - 3y.
- Substitute. α = x - 2(6 - 2x - 3y) = 5x + 6y - 12. Also β = y - 3(6 - 2x - 3y) = 6x + 10y - 18 = 2(3x + 5y - 9).
- So. Putting these into α² + β² = 1 gives (5x + 6y - 12)² + 4(3x + 5y - 9)² = 1, with z = 6 - 2x - 3y. That is option (b).
The idea to keep. For a locus of feet or images, write the new point in terms of the old one, then invert. Put the old point back into its own equation, and the plane equation keeps z in check.
Answer from the NTA final answer key. Solution by PrepWiser.