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

It is given that, (x-a)^2+(y-b)^2=c^2 Differentiating both sides with respect to x, we obtain d dx [(x-a)^2]+ d dx [(y-b)^2]= d dx (c^2) 2(x-a). d dx (x-a)+2(y-b). d dx (y-b)=0 dy dx = -(x-a) y-b (1) d^2y dx^2 = d dx [ -(x-a) y-b ] =- [ (y-b). d dx (x-a)-(x-a). d dx (y-b) (y-b)^2 ] =- [ (y-b)-(x-a). dy dx (y-b)^2 ] =- [ (y-b)-(x-a). -(x-a) y-b (y-b)^2 ] [using (1)] =- [ (y-b)^2+(x-a)^2 (y-b)^3 ] [ 1+ ( dy dx )^2 d^2y dx^2 ]^ 3 2 = [1+ (x-a)^2 (y-b)^2 ]^ 3 2 - [ (y-b)^2+(x-a)^2 (y-b)^3 ] = [ (y)-b)^2+(x-a)^2 (y-b)^2 ]^ 3 2 - [ (y-b)^2+(x-a)^2 (y-b)^3 ] = [ c^2 (y-b)^2 ]^ 3 2 - c^2 (y-b)^3 = c^3 (y-b)^3 - c^2 (y-b)^3 = -c, which is constant and is independent of a and b Hence, proved.

If (x-a)^2+(y-b)^2=c^2, for some c > 0 , prove that [1+ ( dy dx )…

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Question

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

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Answer

It is given that, $\text{(x}-\text{a})^2+(\text{y}-\text{b})^2=\text{c}^2$
Differentiating both sides with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}[(\text{x}-\text{a})^2]+\frac{\text{d}}{\text{dx}}[(\text{y}-\text{b})^2]=\frac{\text{d}}{\text{dx}}(\text{c}^2)$
$\Rightarrow\ 2(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})+2(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})=0$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\ \dots(1)$
$\therefore\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{\text{d}}{\text{dx}}\Big[\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big]$
$=-\Bigg[\frac{(\text{y}-\text{b}).\frac{\text{d}}{\text{dx}}(\text{x}-\text{a})-(\text{x}-\text{a}).\frac{\text{d}}{\text{dx}}(\text{y}-\text{b})}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\frac{\text{dy}}{\text{dx}}}{(\text{y}-\text{b})^2}\Bigg]$
$=-\Bigg[\frac{(\text{y}-\text{b})-(\text{x}-\text{a}).\Big\{\frac{-(\text{x}-\text{a})}{\text{y}-\text{b}}\Big\}}{(\text{y}-\text{b})^2}\Bigg]\ \ [\text{using (1)}]$
$=-\Big[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Big]$
$\therefore\ \Bigg[\frac{1+\Big(\frac{\text{dy}}{\text{dx}}\Big)^2}{\frac{\text{d}^2\text{y}}{\text{dx}^2}}\Bigg]^{\frac{3}{2}}$ $=\frac{\Bigg[1+\frac{(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$ $=\frac{\Bigg[\frac{(\text{y})-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\Bigg[\frac{(\text{y}-\text{b})^2+(\text{x}-\text{a})^2}{(\text{y}-\text{b})^3}\Bigg]}$
$=\frac{\Bigg[\frac{\text{c}^2}{(\text{y}-\text{b})^2}\Bigg]^{\frac{3}{2}}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}=\frac{\frac{\text{c}^3}{(\text{y}-\text{b})^3}}{-\frac{\text{c}^2}{(\text{y}-\text{b})^3}}$
= -c, which is constant and is independent of a and b
Hence, proved.