Let the normals at all the points on a given curve pass through a fixed point (a, b). If the curve passes through (3, -3) and (4, -2sqrt 2 ), and given that a - 2sqrt 2 b = 3, then (a^(2) + b^(2) + ab) is equal to __________.

Let the equation of normal is Y - y = -1 / m(X - x), where, m = dy / dxAs it passes through (a, b)b - y = - 1 / m(a - x) = - dx / dy(a - x) => (b - y)dy = (x - a)dxby - y^2 / 2 = x^2 / 2 - ax + c ..... (i)It passes through (3, -3) (4, -2sqrt 2 ) Therefore - 3b - 9 / 2 = 9 / 2 - 3a + c => - 6b - 9 = 9 - 6a + 2c => 6a - 6b - 2c = 18 => 3a - 3b - c = 9 .... (ii)Also, - 2sqrt 2 b - 4 = 8 - 4a + c4a - 2sqrt 2 b - c = 12 .... (iii)Also, a - 2sqrt 2 b = 3 .... (iv) (given)(ii) - (iii) => - a + ( 2sqrt 2 - 3 )b = - 3 ... (v)(iv) + (v) => b = 0,a = 3 Therefore a^2 + b^2 + ab = 9

Let the normals at all the points on a given curve pass... | null (Mathematics)

Let the equation of normal is Y

-

y =

-

{1 \over m}(X - x)

, where, m =

{{dy} \over {dx}}

As it passes through (a, b)

b - y = - {1 \over m}(a - x) = - {{dx} \over {dy}}(a - x)

\Rightarrow (b - y)dy = (x - a)dx

- {{{y^2}} \over 2} = {{{x^2}} \over 2} - ax + c

..... (i)

It passes through (3,

-

3) & (4,

-

\sqrt 2

)

\therefore

- 3b - {9 \over 2} = {9 \over 2} - 3a + c

\Rightarrow - 6b - 9 = 9 - 6a + 2c

\Rightarrow 6a - 6b - 2c = 18

\Rightarrow 3a - 3b - c = 9

.... (ii)

Also,

- 2\sqrt 2 b - 4 = 8 - 4a + c

4a - 2\sqrt 2 b - c = 12

.... (iii)

Also,

a - 2\sqrt 2 \,b = 3

.... (iv) (given)

(ii) - (iii) \Rightarrow - a + \left( {2\sqrt 2 - 3} \right)b = - 3

... (v)

(iv) + (v) \Rightarrow b = 0,a = 3

\therefore

{a^2} + {b^2} + ab = 9

Explanation

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