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Hint: Use the quotient rule of differentiation in the given question to get the desired result.

Here, starting with \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\], we have to prove that \[\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}\].

Also, we have to prove \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\]for parabola \[{{y}^{2}}=4ax\].

Taking, \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

Now, we will differentiate both sides with respect to \[y\].

Also, we know that quotient rule says that,

\[\dfrac{d}{dy}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{dt}{dy} \right)-f\left( \dfrac{dg}{dy} \right)}{{{g}^{2}}}\]

In \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

\[f=1\]and \[g=\dfrac{dy}{dx}\]

Now, differentiating both sides with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{\left( \dfrac{dy}{dx} \right)\dfrac{d}{dy}\left( 1 \right)-\left( 1 \right)\left[ \dfrac{d}{dy}\left( \dfrac{dy}{dx} \right) \right]}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

As we know that \[\dfrac{d}{dy}\left( \text{constant} \right)=0\]

Therefore, we get

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{0-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

Now we will multiply by \[\dfrac{dy}{dx}\] on both the numerator and denominator of \[RHS\].

We get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}.\dfrac{dy}{dx}}\]

We know that \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Therefore, we get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\][Hence Proved]

which is our required result.

Now, we have to prove that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\] for parabola \[{{y}^{2}}=4ax\].

Now we take parabola, \[{{y}^{2}}=4ax\].

So, we differentiate the above equation with respect to \[x\].

Also, we know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Therefore, \[2y\dfrac{dy}{dx}=4a\]

\[\dfrac{dy}{dx}=\dfrac{2a}{y}....\left( i \right)\]

Again differentiating both sides with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( 2a \right){{y}^{-1-1}}.\dfrac{dy}{dx}\]

Now, we put the value of \[\dfrac{dy}{dx}\]from equation \[\left( i \right)\].

We get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2a}{{{y}^{2}}}.\dfrac{2a}{y}\]

Therefore, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4{{a}^{2}}}{{{y}^{3}}}....\left( ii \right)\]

Now, from previous results, we know that

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{-{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

By putting values from equation \[\left( i \right)\]and \[\left( ii \right)\]

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\left[ \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right]}{{{\left[ \dfrac{2a}{y} \right]}^{3}}}\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{+4{{a}^{2}}}{{{y}^{3}}}}{\dfrac{8{{a}^{3}}}{{{y}^{3}}}}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{+1}{2a}....\left( iii \right)\]

Now we will multiply the equation \[\left( ii \right)\]and \[\left( iii \right)\].

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\left( \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right).\dfrac{1}{2a}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-2a}{{{y}^{3}}}\]which is the required result.

Note: In the term \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)\], some students cancel \[dy\] from numerator and denominator considering them to be like terms but that is wrong and \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)=\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{\dfrac{dy}{dx}}\].

Here, starting with \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\], we have to prove that \[\dfrac{{{d}^{2}}y}{d{{y}^{2}}}=-\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{3}}}\].

Also, we have to prove \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\]for parabola \[{{y}^{2}}=4ax\].

Taking, \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

Now, we will differentiate both sides with respect to \[y\].

Also, we know that quotient rule says that,

\[\dfrac{d}{dy}\left( \dfrac{f}{g} \right)=\dfrac{g\left( \dfrac{dt}{dy} \right)-f\left( \dfrac{dg}{dy} \right)}{{{g}^{2}}}\]

In \[\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}\]

\[f=1\]and \[g=\dfrac{dy}{dx}\]

Now, differentiating both sides with respect to \[y\].

We get, \[\dfrac{d}{dy}\left( \dfrac{dx}{dy} \right)=\dfrac{\left( \dfrac{dy}{dx} \right)\dfrac{d}{dy}\left( 1 \right)-\left( 1 \right)\left[ \dfrac{d}{dy}\left( \dfrac{dy}{dx} \right) \right]}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

As we know that \[\dfrac{d}{dy}\left( \text{constant} \right)=0\]

Therefore, we get

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{0-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

Now we will multiply by \[\dfrac{dy}{dx}\] on both the numerator and denominator of \[RHS\].

We get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}.\dfrac{dy}{dx}}\]

We know that \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right).\dfrac{dy}{dx}=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\]

Therefore, we get \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\][Hence Proved]

which is our required result.

Now, we have to prove that \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=-\dfrac{2a}{{{y}^{3}}}\] for parabola \[{{y}^{2}}=4ax\].

Now we take parabola, \[{{y}^{2}}=4ax\].

So, we differentiate the above equation with respect to \[x\].

Also, we know that \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\]

Therefore, \[2y\dfrac{dy}{dx}=4a\]

\[\dfrac{dy}{dx}=\dfrac{2a}{y}....\left( i \right)\]

Again differentiating both sides with respect to \[x\],

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( 2a \right){{y}^{-1-1}}.\dfrac{dy}{dx}\]

Now, we put the value of \[\dfrac{dy}{dx}\]from equation \[\left( i \right)\].

We get \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2a}{{{y}^{2}}}.\dfrac{2a}{y}\]

Therefore, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-4{{a}^{2}}}{{{y}^{3}}}....\left( ii \right)\]

Now, from previous results, we know that

\[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{-{{d}^{2}}y}{d{{x}^{2}}}}{{{\left( \dfrac{dy}{dx} \right)}^{2}}}\]

By putting values from equation \[\left( i \right)\]and \[\left( ii \right)\]

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-\left[ \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right]}{{{\left[ \dfrac{2a}{y} \right]}^{3}}}\]

\[\Rightarrow \dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{\dfrac{+4{{a}^{2}}}{{{y}^{3}}}}{\dfrac{8{{a}^{3}}}{{{y}^{3}}}}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{+1}{2a}....\left( iii \right)\]

Now we will multiply the equation \[\left( ii \right)\]and \[\left( iii \right)\].

We get, \[\left( \dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\left( \dfrac{{{d}^{2}}x}{d{{y}^{2}}} \right)=\left( \dfrac{-4{{a}^{2}}}{{{y}^{3}}} \right).\dfrac{1}{2a}\]

By cancelling the like terms,

We get, \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}.\dfrac{{{d}^{2}}x}{d{{y}^{2}}}=\dfrac{-2a}{{{y}^{3}}}\]which is the required result.

Note: In the term \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)\], some students cancel \[dy\] from numerator and denominator considering them to be like terms but that is wrong and \[\dfrac{d}{dy}\left( \dfrac{dy}{dx} \right)=\dfrac{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}{\dfrac{dy}{dx}}\].

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