Question

The foot of the perpendicular from a point on the circle $x ^2+ y ^2=1, z =0$ 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 1

Answer: (B)

$(5 x+6 y-12)^2+4(3 x+5 y-9)^2=1$, $z=6-2 x-3 y$

Answer 2

The correct option is (B): $(5 x+6 y-12)^2+4(3 x+5 y-9)^2=1$, $z=6-2 x-3 y$