The differential equation of the family of hy | vedclass

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

Question

(A) The equation of the family of hyperbolas with centres at the origin and axes along the coordinate axes is given by $\frac{x^2}{a^2} - \frac{y^2}{b

Vedclass · Accepted Answer

$x y y_2 + x y_1^2 - y y_1 = 0$

Vedclass · Answer

(A) The equation of the family of hyperbolas with centres at the origin and axes along the coordinate axes is given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ (or $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for the general conic form). Differentiating with respect to $x$: $\frac{2x}{a^2} - \frac{2y}{b^2} y_1 = 0 \Rightarrow \frac{x}{a^2} = \frac{y y_1}{b^2} \Rightarrow \frac{y y_1}{x} = \frac{b^2}{a^2} = k$ (constant).Differentiating again with respect to $x$: $\frac{d}{dx} \left( \frac{y y_1}{x} ight) = 0$.Using the quotient rule: $\frac{x(y y_2 + y_1^2) - y y_1}{x^2} = 0$.Since $x 
eq 0$,we have $x y y_2 + x y_1^2 - y y_1 = 0$.

The differential equation of the family of hyperbolas having their centres at the origin and their axes along the coordinate axes is

Similar Questions

The differential equation of the family of all parabolas whose axis is the $y$-axis is ...

If $m$ and $n$ are respectively the order and the degree of the differential equation representing the family of curves $y^2 - 5ax - 5a^{3/2} = 0$ ($a > 0$ is a parameter), then the value of $m - n$ is

The differential equation corresponding to the family of circles given by $(x-a)^2+(y-b)^2=4$,where $a$ and $b$ are parameters,is

The differential equation of the family of curves $y = a e^x + b x e^x + c x^2 e^x$,where $a, b, c$ are arbitrary constants,is

Form the differential equation of the family of circles in the first quadrant which touch the coordinate axes.

Vedclass Test Series

Exam Paper Generator

Online Exam Module