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Question: Find the equation of the tangent plane and normal to the surface \(2{{x}^{2}}+{{y}^{2}}+2z=3\) at th...

Find the equation of the tangent plane and normal to the surface 2x2+y2+2z=32{{x}^{2}}+{{y}^{2}}+2z=3 at the point (2,1,−3)\left( 2,1,-3 \right).

Explanation

Solution

Hint: For this type of question we know that the equation of tangent plane at a point is given by the formula: z=fx(x0,y0,z0)(x−x0)+fy(x0,y0,z0)(y−y0)+fz(x0,y0,z0)(z−z0)=0z={{f}_{x}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( x-{{x}_{0}} \right)+{{f}_{y}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( y-{{y}_{0}} \right)+{{f}_{z}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( z-{{z}_{0}} \right)=0 and equation of normal plane at any point is given by: x−x0fx(x0,y0,z0)=y−y0fy(x0,y0,z0)=z−z0fz(x0,y0,z0)\dfrac{x-{{x}_{0}}}{{{f}_{x}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)}=\dfrac{y-{{y}_{0}}}{{{f}_{y}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)}=\dfrac{z-{{z}_{0}}}{{{f}_{z}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)} Using these formulas, we can find the equation of tangent plane and normal to the surface.

Complete step-by-step answer:
Now, as given in the question, we will consider 2x2+y2+2z=32{{x}^{2}}+{{y}^{2}}+2z=3as ϕ(x0,y0,z0)\phi \left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right) and (2,1,−3)\left( 2,1,-3 \right) as point P.
Before moving further we will understand what a tangent plane is? In three-dimensional space, many lines can behave as a tangent to a point, so if these lines stay in the same plane then that plane is called a tangent plane.
Now, we know that the equation of tangent to surface at any plane at a point is given by the formula: z=fx(x0,y0,z0)(x−x0)+fy(x0,y0,z0)(y−y0)+fz(x0,y0,z0)(z−z0)=0z={{f}_{x}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( x-{{x}_{0}} \right)+{{f}_{y}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( y-{{y}_{0}} \right)+{{f}_{z}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( z-{{z}_{0}} \right)=0
Now, taking the equation of surface as, 2x2+y2+2z=32{{x}^{2}}+{{y}^{2}}+2z=3 and taking everything to the left-hand side we get,
2x2+y2+2z−3=02{{x}^{2}}+{{y}^{2}}+2z-3=0 …………………………….(i)
Consider the expression (i) as ϕ(x,y,z)=2x2+y2+2z−3\phi \left( x,y,z \right)=2{{x}^{2}}+{{y}^{2}}+2z-3, now performing partial derivation to each component of the equation we will get,
∂ϕ∂x=∂∂x(2x2+y2+2z−3)=4x\dfrac{\partial \phi }{\partial x}=\dfrac{\partial }{\partial x}\left( 2{{x}^{2}}+{{y}^{2}}+2z-3 \right)=4x
In partial derivation of an equation with respect to x, all the variables except x become zero and only derivation of x is considered. Similarly, we will do this for y and z components with respect to y and z respectively.
∂ϕ∂y=∂∂y(2x2+y2+2z−3)=2y\dfrac{\partial \phi }{\partial y}=\dfrac{\partial }{\partial y}\left( 2{{x}^{2}}+{{y}^{2}}+2z-3 \right)=2y
∂ϕ∂z=∂∂z(2x2+y2+2z−3)=2\dfrac{\partial \phi }{\partial z}=\dfrac{\partial }{\partial z}\left( 2{{x}^{2}}+{{y}^{2}}+2z-3 \right)=2.
Now, we will consider that these components ∂ϕ∂x,∂ϕ∂y and ∂ϕ∂z\dfrac{\partial \phi }{\partial x},\dfrac{\partial \phi }{\partial y}\ \text{and }\dfrac{\partial \phi }{\partial z}are at point P, so, we will substitute the value of point P in x, y and z components respectively and by doing this we will get,
∂ϕ∂x=4×2=8\dfrac{\partial \phi }{\partial x}=4\times 2=8
∂ϕ∂y=2×1=2\dfrac{\partial \phi }{\partial y}=2\times 1=2
∂ϕ∂z=2\dfrac{\partial \phi }{\partial z}=2
Now, we will substitute the values in equation z=fx(x0,y0,z0)(x−x0)+fy(x0,y0,z0)(y−y0)+fz(x0,y0,z0)(z−z0)=0z={{f}_{x}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( x-{{x}_{0}} \right)+{{f}_{y}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( y-{{y}_{0}} \right)+{{f}_{z}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)\left( z-{{z}_{0}} \right)=0,
where fx(x0,y0,z0)=∂ϕ∂x,fy(x0,y0,z0)=∂ϕ∂y and fz(x0,y0,z0)=∂ϕ∂z{{f}_{x}}\left( {{x}_{0,}}{{y}_{0,}}{{z}_{0}} \right)=\dfrac{\partial \phi }{\partial x},{{f}_{y}}\left( {{x}_{0,}}{{y}_{0,}}{{z}_{0}} \right)=\dfrac{\partial \phi }{\partial y}\ \text{and }{{f}_{z}}\left( {{x}_{0,}}{{y}_{0,}}{{z}_{0}} \right)=\dfrac{\partial \phi }{\partial z} and x0=2,y0=1 and z0=−3{{x}_{0}}=2,{{y}_{0}}=1\ \text{and }{{\text{z}}_{0}}=-3.
Substituting the values in equation we will get,
(x−2)8+(y−1)2+(z+3)2=0\left( x-2 \right)8+\left( y-1 \right)2+\left( z+3 \right)2=0
⇒4x+y+z=6\Rightarrow 4x+y+z=6
Hence, the equation of tangent plane to the surface is 4x+y+z=64x+y+z=6.
Now, we know that equation of normal to the surface at any point is given by:
x−x0fx(x0,y0,z0)=y−y0fy(x0,y0,z0)=z−z0fz(x0,y0,z0)\dfrac{x-{{x}_{0}}}{{{f}_{x}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)}=\dfrac{y-{{y}_{0}}}{{{f}_{y}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)}=\dfrac{z-{{z}_{0}}}{{{f}_{z}}\left( {{x}_{0}},{{y}_{0}},{{z}_{0}} \right)}.
Again, substituting the values in the equation, we will get,
x−28=y−12=z+32\dfrac{x-2}{8}=\dfrac{y-1}{2}=\dfrac{z+3}{2}
⇒x−24=y−1=z+3\Rightarrow \dfrac{x-2}{4}=y-1=z+3.
Hence, the equation of normal to the surface at any point is given by, x−24=y−1=z+3\dfrac{x-2}{4}=y-1=z+3.

Note: While solving such types of sums students must know the formula of tangent plane and normal plane to the surface at any point. Students must take care while using the formula as they might use the formula of tangent plane in normal plane or vice-versa. Students should also take care while substituting the values in equations as they might substitute wrong values and due to that final answer may differ.