Question

The foot of the perpendicular from a point on the circle $x^2+y^2=1,z=0$ to the plane $2x+3y+z=6$ lies on which one of the following curves ?

• A. $(6x+5y-12{)}^2+4(3x+7y-8{)}^2=1$, $z=6-2x-3y$
• B. $(5x+6y-12{)}^2+4(3x+5y-9{)}^2=1$, $z=6-2x-3y$
• C. $(6x+5y-14{)}^2+9(3x+5y-7{)}^2=1$, $z=6-2x-3y$
• D. $(5x+6y-14{)}^2+9(3x+7y-8{)}^2=1$, $z=6-2x-3y$

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 $\cos \theta$
$(5h+6k-12{)}^2+4(3h+5k-9{)}^2=1$