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)

$ \frac{ h -\cos \theta}{2}=\frac{ k -\sin \theta}{3}=\frac{ w -0}{1} $
$ =\frac{-1(2 \cos \theta+3 \sin \theta-6)}{14} $
$ h =\cos \frac{-2(2 \cos \theta+3 \sin \theta-6)}{14} $
$ =\frac{10 \cos \theta-6 \sin \theta+12}{14} $
$ k =\sin \theta-\frac{3}{14}(2 \cos \theta+3 \sin \theta-6) $
$ k =\frac{5 \sin \theta-6 \cos \theta+18}{14}$
Elementary $\sin \theta$ and $\cos \theta$
$(5 h+6 k-12)^2+4(3 h+5 k-9)^2=1$