If frac{x^2}{a^2}+frac{y^2}{b^2}=1, then frac | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Differentiating both sides with respect to $x$:$\frac{2x}{a^2} + \frac{

Vedclass · Accepted Answer

$-\frac{b^4}{a^2 y^3}$

Vedclass · Answer

(A) Given the equation of the ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Differentiating both sides with respect to $x$:$\frac{2x}{a^2} + \frac{2y}{b^2} \cdot \frac{dy}{dx} = 0$$\frac{y}{b^2} \cdot \frac{dy}{dx} = -\frac{x}{a^2}$$\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$Now, differentiate again with respect to $x$ using the quotient rule:$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \frac{d}{dx} \left( \frac{x}{y} \right)$$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y(1) - x(\frac{dy}{dx})}{y^2} \right)$Substitute $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$:$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y - x(-\frac{b^2 x}{a^2 y})}{y^2} \right)$$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{y + \frac{b^2 x^2}{a^2 y}}{y^2} \right) = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 y^2 + b^2 x^2}{a^2 y^3} \right)$Since $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $b^2 x^2 + a^2 y^2 = a^2 b^2$.$\frac{d^2 y}{dx^2} = -\frac{b^2}{a^2} \cdot \left( \frac{a^2 b^2}{a^2 y^3} \right) = -\frac{b^4}{a^2 y^3}$.

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

Similar Questions

If $f(x)=x^3-x^2 f^{\prime}(1)+x f^{\prime \prime}(2)-f^{\prime \prime \prime}(3), x \in R$,then

The equation $x^5 - 5x^3 + 5x^2 - 1 = 0$ has how many equal roots?

If $y = 100 e^{2x} + 200 e^{-2x}$ and $\frac{d^2 y}{dx^2} = ay$,then $a = $ . . . . . .

If ${I_n} = \frac{d^n}{dx^n}(x^n \log x)$,then ${I_n} - n{I_{n - 1}} = $

Find the second order derivative of the function $y = e^{x} \sin 5x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module