Question

If $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, then $\frac{d^{2}y}{d{x}^2}=$

• A. $\frac{-b^4}{a^2{y}^3}$
• B. $\frac{b^2}{a{y}^2}$
• C. $\frac{{-b}^3}{a^2{y}^3}$
• D. $\frac{b^3}{a^2{y}^2}$

Answer 1

The correct option is A $\frac{-b^4}{a^2{y}^3}$

Given, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow b^2{x}^2+a^2{y}^2=a^2{b}^2$

On differentiating, we get
$2a^{2}y{y}_1=-2b^{2}x$
$\Rightarrow a^{2}y{y}_1=-b^{2}x$

Again differentiating, we get

$a^2[y\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2]=-b^2$

$\Rightarrow a^2[y\frac{d^{2}y}{d{x}^2}+\frac{b^4{x}^2}{y^2{a}^4}]=-b^2$

Thus we have : $[y\frac{d^{2}y}{d{x}^2}+\frac{b^4{x}^2}{y^2{a}^4}]=\frac{-b^2}{a^2}$

Substituting $\frac{x^2}{a^2}=1-\frac{y^2}{b^2}$ in the above equation we have,

$y\frac{d^{2}y}{d{x}^2}+\frac{(b^2-y^2)\times b^2}{a^2{y}^2}=\frac{-b^2}{a^2}$

$\Rightarrow \frac{d^{2}y}{d{x}^2}=\frac{-b^2}{a^{2}y}-\frac{(b^2-y^2)\times b^2}{a^2{y}^3}$

Thus, $\frac{d^{2}y}{d{x}^2}=\frac{-b^2}{a^{2}y}-\frac{b^2\times b^2}{a^2{y}^3}+\frac{-b^2}{a^{2}y}$

$\Rightarrow$ $\frac{d^{2}y}{d{x}^2}=\frac{-b^4}{a^2{y}^3}$