X 2 a 2 y 2 b 2 1

Asked by admin @ 11/08/2021 in Math viewed by 331 People

If x^2/a^2+y^2/b^2=1. Prove that d^2y/dx^2=-b^4/a^2y^3

Answered by admin @ 11/08/2021

Given

$\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

To Prove

$\rm \dfrac{d^2y}{dx^2}=-\dfrac{b^2}{a^2y^3}$

Solution

Note :

y' = first derivative

y" = second derivative

Given ,

$\rm \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

Differentiate with respect to x on both sides ,

$\to \rm \dfrac{2x}{a^2}+\dfrac{2yy'}{b^2}=0\\\\\to \rm \dfrac{x}{a^2}+\dfrac{yy'}{b^2}=0...(1)\\\\\to \rm \dfrac{y'}{b^2}=-\dfrac{x}{a^2y}\\\\\to \rm \dfrac{(y')^2}{b^4}=\dfrac{x^2}{a^4y^2}\\\\\to \rm \dfrac{(y')^2}{b^2}=\dfrac{x^2b^2}{a^4y^2}...(2)$

Differentiate (1) again ,

$\to \rm \dfrac{1}{a^2}+\dfrac{(y')^2}{b^2}+\dfrac{yy"}{b^2}=0\\\\\to \rm \dfrac{1}{a^2}+\dfrac{x^2b^2}{a^4y^2}+\dfrac{yy"}{b^2}=0\ [\bf From\ (2)]\\\\\to \rm \dfrac{1}{a^2}\left( 1+\dfrac{x^2}{a^2}.\dfrac{b^2}{y^2}\right)+\dfrac{yy"}{b^2}=0\\\\\to \rm \dfrac{1}{a^2}\left( 1+\left[ 1-\dfrac{y^2}{b^2}\right]\dfrac{b^2}{y^2}\right)+\dfrac{yy"}{b^2}=0\\\\\to \rm \dfrac{1}{a^2}\left(1+\dfrac{b^2}{y^2}-1\right)+\dfrac{yy"}{b^2}=0\\\\\to \rm \dfrac{b^2}{a^2y^2}+\dfrac{yy"}{b^2}=0\\\\$

$\to \rm \dfrac{y"y}{b^2}=-\dfrac{b^2}{a^2y^2}\\\\\to \rm y"=-\dfrac{b^4}{a^2y^3}\\\\\to \bf \dfrac{d^2x}{dx^2}=-\dfrac{b^4}{a^2y^3}$

Hence proved

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