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 $y = \frac{A}{x} + B x^2$, then $x^2 \frac{d^2 y}{d x^2} =$

If $y=(\tan^{-1} x)^{2}$,show that $(x^{2}+1)^{2} y_{2}+2 x(x^{2}+1) y_{1}=2$.

If $y=A \cos n x+B \sin n x$,then $y_2+n^2 y$ is equal to

Let $f(x)$ be a polynomial in $x$. Then the second derivative of $f(e^x)$ is:

If $f: R \rightarrow R$ is an even function which is twice differentiable on $R$ and $f^{\prime \prime}(\pi)=1$, then $f^{\prime \prime}(-\pi)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module