Challenging +1.8 This is a Further Maths question requiring implicit differentiation to find the normal vector to the surface, then using the angle condition between planes (involving dot products and the angle formula) to derive a constraint. While it involves multiple concepts (3D geometry, partial derivatives, plane angles), the techniques are standard for Further Pure. The 6-mark allocation and systematic approach needed place it well above average difficulty but not at the extreme end.
A surface \(S\) is defined for \(z \geqslant 0\) by \(x^2 + y^2 + 2z^2 = 126\). \(C\) is the set of points on \(S\) for which the tangent plane to \(S\) at that point intersects the \(x\)-\(y\) plane at an angle of \(\frac{1}{4}\pi\) radians.
Show that \(C\) lies in a plane, \(\Pi\), whose equation should be determined. [6]
Question 5:
5 | ∂g ∂g ∂g
=2x or =2yor =4z
∂x ∂y ∂z
2x
∇g = 2y is the normal to the tangent plane at
4z
each point.
2x 0
∇g.n= 2y . 0 =4z
4z 1
2x 0
π
= 2y 0 cos
3
4z 1
1
= (2x)2+(2y)2+(4z)2 ×1×
2
1
=2 x2+ y2+4z2 × = 126−2z2+4z2
2
= 126+2z2
126+2z2 =4z⇒126+2z2 =16z2
⇒14z2 =126⇒z2 =9⇒z=±3
z≥0⇒z=3 which is the equation of Π. | M1
A1
M1
M1
M1
A1
[6] | 3.1a
3.1a
3.1a
2.2a
1.1
3.2a | g(x, y, z) = x2 + y2 + 2z2 and
surface is g = 126. Finding one
correct partial derivative.
Finding the normal vector.
Dotting normal with normal to x-y
plane.
Expressing dot product in other
form using correct value of angle.
π 1
Using cos = , forming
3 2
magnitude of both normals and
reducing to form a+bz2 oe
(could be done after squaring).
Not ± in final answer. | May be rewritten as
z=f(x, y)= 63−1 y2−1x2
2 2
but condone ±.
x
−
2z
y
∇g = − oe
2z
−1
or a+bx2+by2 (could see eg
x2 + y2 =108 oe after equating to
4z and eliminating z).
A surface $S$ is defined for $z \geqslant 0$ by $x^2 + y^2 + 2z^2 = 126$. $C$ is the set of points on $S$ for which the tangent plane to $S$ at that point intersects the $x$-$y$ plane at an angle of $\frac{1}{4}\pi$ radians.
Show that $C$ lies in a plane, $\Pi$, whose equation should be determined. [6]
\hfill \mbox{\textit{OCR MEI Further Extra Pure 2021 Q5 [6]}}