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

CBSE Class 12 · 2019 · Differential EquationsLong Answer

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

, for some

c > 0

, prove that

\frac{[ 1 + ( \frac{d y}{d x} ) ^{2} ] ^{3/2}}{\frac{d ^{2} y}{d x ^{2}}}

is a constant independent of

a

and

b

Answer: Proved

Differentiate

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

(x - a) + (y - b) y^{'} = 0 \Rightarrow y^{'} = - \frac{x - a}{y - b}

Differentiate again and simplify using the relation:

y^{''} = - \frac{( y - b ) ^{2} + ( x - a ) ^{2}}{( y - b ) ^{3}} = - \frac{c ^{2}}{( y - b ) ^{3}}

Now

\frac{[ 1 + ( y ^{'} ) ^{2} ] ^{3/2}}{y ^{''}} = \frac{[ 1 + \frac{( x - a ) ^{2}}{( y - b ) ^{2}} ] ^{3/2}}{- \frac{c ^{2}}{( y - b ) ^{3}}} = \frac{( \frac{c ^{2}}{( y - b ) ^{2}} ) ^{3/2} ( y - b ) ^{3}}{- c ^{2}} = - c

which is constant (independent of

a, b

Proved; the constant is - c

CBSE Class 12 Board · 2019 · Subjective