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Joint Entrance Examination
Mathematics
3D Geometry
Question 492
Question 492 of 521
Medium
MCQ
The foot of the perpendicular from a point on the circle
x
2
+
y
2
=
1
,
z
=
0
x^{2}+y^{2}=1, z=0
x
2
+
y
2
=
1
,
z
=
0
to the plane
2
x
+
3
y
+
z
=
6
2 x+3 y+z=6
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
\[(5 x+6 y-14)^{2}+9(3 x+7 y-8)^{2}=1, z=6-2 x-3 y\]
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Official solution
Any point on
x
2
+
y
2
=
1
{x^2} + {y^2} = 1
x
2
+
y
2
=
1
,
z
=
0
z = 0
z
=
0
is \[p(\cos \theta ,\,\sin \theta ,\,0)\] If foot of perpendicular of p on the plane
2
x
+
3
y
+
z
=
6
2x + 3y + z = 6
2
x
+
3
y
+
z
=
6
is
(
h
,
k
,
l
)
(h,k,l)
(
h
,
k
,
l
)
then \[{{h - \cos \theta } \over 2} = {{k - \sin \theta } \over 3} = {{l - 0} \over 1}\] \[ = - \left( {{{2\cos \theta + 3\sin \theta + 0 - 6} \over {{2^2} + {3^2} + {1^2}}}} \right) = r\] (let) \[h = 2r + \cos \theta ,\,k = 3r + \sin \theta ,\,l = r\] Hence, \[h - 2l = \cos \theta \] and \[k - 3l = \sin \theta \] Hence,
(
h
−
2
l
)
2
+
(
k
−
3
l
)
2
=
1
{(h - 2l)^2} + {(k - 3l)^2} = 1
(
h
−
2
l
)
2
+
(
k
−
3
l
)
2
=
1
When
l
=
6
−
2
h
−
3
k
l = 6 - 2h - 3k
l
=
6
−
2
h
−
3
k
Hence required locus is \[{(x - 2(6 - 2x - 3y))^2} + {(y - 3(6 - 2x - 3y))^2} = 1\] \[ \Rightarrow {(5x + 6y - 12)^2} + 4{(3x + 5y - 9)^2} = 1,\,z = 6 - 2x - 3y\]
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