Math, asked by Makwanasunil4710, 10 months ago

If (x – a)² + (y – b)² = c², for some c > 0, prove that \frac{\Big [1+\Big(\frac{dy}{dx}\Big)^{2} \Big]^{\frac{3}{2} }  }{\frac{d^{2}y }{dx^{2} } } is a constant independent of a and b.

Answers

Answered by Anonymous
13

AnswEr :

Given,

 \sf \:  {(x - a)}^{2}  +  {(y -b )}^{2}  =  {c}^{2}

Differentiating w.r.t x on both sides,

 \implies \sf \:  2{(x - a)} +  2{(y -b )}   \times  \dfrac{d(y - b)}{dx} =  0 \\  \\  \implies \sf \: 2(x - a) =  - 2(y - b) \dfrac{dy}{dx}  \\   \\  \implies \boxed{ \boxed{ \sf \:  \dfrac{dy}{dx}  =   - \dfrac{x - a}{y - b} }}

Differentiating y' w.r.t x, we get :

 \sf \dfrac{ {d}^{2}y }{  {dx}^{2} }  =   -  \dfrac{   \dfrac{d(x - a)}{dx}(y - b)   -   \dfrac{d(y - b)}{dx}(x - a) }{(y - b) {}^{2} }  \\  \\  \implies \sf \:  \dfrac{ {d}^{2}y }{dx {}^{2} }  =   - \dfrac{(y - b) - (x - a) \dfrac{dy}{dx} }{(y - b) {}^{2} }

Substituting value of dy/dx in the above equation :

 \sf \implies \:  \dfrac{ {d}^{2} y}{ {dx}^{2} }  =  -  \dfrac{(y - b) {}^{2}  + (x - a) {}^{2} }{(y - b) {}^{3} }  \\  \\  \implies \sf \: \dfrac{ {d}^{2} y}{ {dx}^{2} } =  \frac{ - c {}^{2} }{(y - b) {}^{3} }

Now,

 \sf \dfrac{\Big [1+\Big(\dfrac{dy}{dx}\Big)^{2} \Big]^{\frac{3}{2} } }{\dfrac{d^{2}y }{dx^{2} } } \\  \\  \longrightarrow \:  \sf \:   \dfrac{\bigg[1 +  \dfrac{(x - a) {}^{2} }{(y - b) {}^{2} }  \bigg]  {}^{ \frac{3}{2} }  }{ \dfrac{ -  {c}^{2} }{(y - b) {}^{3} } } \\  \\  \longrightarrow \sf \: \dfrac{\bigg[ \dfrac{(x - a) {}^{2} + (y - b) {}^{2} }{(y - b) {}^{2} }  \bigg]  {}^{ \frac{3}{2} }  }{ \dfrac{ -  {c}^{2} }{(y - b) {}^{3} } } \\   \\

 \longrightarrow \sf \:  \dfrac{  \bigg(  \dfrac{c {}^{2} }{{(y - b)} {}^{2} } \bigg) {}^{ \frac{3}{2} }  }{ \dfrac{ - c {}^{2} }{(y - b) {}^{3} } }  \\  \\  \longrightarrow \:  \sf \:  -  \dfrac{c {}^{3} }{ {c}^{2} }  \times  \dfrac{(y - b) {}^{3} }{(y - b) {}^{3} }  \\  \\  \longrightarrow \sf \:  - c

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