"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, $-$2$\sqrt 2$), and given that a $-$ 2$\sqrt 2$ b = 3, then (a2 + b2 + ab) is equal to __________."

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Accepted Answer

"Let the equation of normal is Y $-$ y = $-$${1 \over m}(X - x)$, where, m = ${{dy} \over {dx}}$

As it passes through (a, b)

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

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

by $- {{{y^2}} \over 2} = {{{x^2}} \over 2} - ax + c$ ..... (i)

It passes through (3, $-$3) & (4, $-$2$\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)

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

$\therefore$ ${a^2} + {b^2} + ab = 9$"

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, $-$2$\sqrt 2$), and given that a $-$ 2$\sqrt 2$ b = 3,
then (a² + b² + ab) is equal to __________.

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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, $-$2$\sqrt 2$), and given that a $-$ 2$\sqrt 2$ b = 3, then (a2 + b2 + ab) is equal to __________.

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More Application of Derivatives questions

Drill 25 more like these. Every day. Free.

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, $-$2$\sqrt 2$), and given that a $-$ 2$\sqrt 2$ b = 3,
then (a² + b² + ab) is equal to __________.