Question

A
The equation of the normal to the surface 2x2 + y2 + 22 = 3 at (2, 1, -3) is

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Answer 1

$\large\underline{\sf{Solution-}}$

We know that,

• If f(x, y, z) be any surface, then equation of normal to the surface at (a, b, c) is given by

$\: \: \: \: \: \: \boxed{ \sf \: \dfrac{x - a}{ \: \dfrac{\partial \:f}{\partial \:x} \: } = \dfrac{y - b}{\dfrac{\partial \:f}{\partial \:y} } = \dfrac{z - c}{\dfrac{\partial \:f}{\partial \:z} } }$

Now,

Given surface is

$\rm :\longmapsto\:f = {2x}^{2} + {y}^{2} + 2z - 3 = 0 - - - (1)$

Differentiate partially w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{\partial \:f}{\partial \:x} = 4x$

Therefore,

$\rm :\implies\: \: \bigg(\dfrac{\partial \:f}{\partial \:x} \bigg)_{(2,1,-3)} = 4 \times 2 = 8$

Now,

Again Differentiate (1) partially w. r. t. y, we get

$\rm :\longmapsto\:\dfrac{\partial \:f}{\partial \:y} = 2y$

Therefore,

$\rm :\implies\: \: \bigg(\dfrac{\partial \:f}{\partial \:y} \bigg)_{(2,1,-3)} = 2 \times 1 = 2$

Now,

Again differentiate (1) partially w. r. t. z, we get

$\rm :\longmapsto\:\dfrac{\partial \:f}{\partial \:z} = 2$

Now, we have

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: \bigg(\dfrac{\partial \:f}{\partial \:x} \bigg)_{(2,1,-3)} = 8$

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: \bigg(\dfrac{\partial \:f}{\partial \:y} \bigg)_{(2,1,-3)} = 2$

$\: \: \: \: \: \: \: \: \bull \: \sf \: \: \: \bigg(\dfrac{\partial \:f}{\partial \:z} \bigg)_{(2,1,-3)} = 2$

So, equation of normal to the surface at the point (2, 1, - 3) is

$\rm :\longmapsto\:{ \sf \: \dfrac{x - a}{ \: \dfrac{\partial \:f}{\partial \:x} \: } = \dfrac{y - b}{\dfrac{\partial \:f}{\partial \:y} } = \dfrac{z - c}{\dfrac{\partial \:f}{\partial \:z} } }$

On substituting the values, we get

$\rm :\longmapsto\:\dfrac{x - 2}{8} = \dfrac{y - 1}{2} = \dfrac{z + 3}{2}$

$\rm :\longmapsto\:\dfrac{x - 2}{4} = \dfrac{y - 1}{1} = \dfrac{z + 3}{1}$

Additional Information :-

If f(x, y, z) be any surface, then equation of tangent plane to the surface at (a, b, c) is given by

$\boxed{ \sf \: (x - a)\dfrac{\partial \:f}{\partial \:x} \: + \: (y - b)\dfrac{\partial \:f}{\partial \:y} \: + \: (z - c)\dfrac{\partial \:f}{\partial \:z} = 0}$