A differential equation representing the fami | vedclass

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

Question

(D) The length of the latus rectum $4a$ is the distance from the point $(2, -3)$ to the line $3x + 4y - 5 = 0$.$4a = \frac{|3(2) + 4(-3) - 5|}{\sqrt{3

Vedclass · Accepted Answer

$11 \frac{d^{2}y}{dx^{2}} = 10$

Vedclass · Answer

(D) The length of the latus rectum $4a$ is the distance from the point $(2, -3)$ to the line $3x + 4y - 5 = 0$.$4a = \frac{|3(2) + 4(-3) - 5|}{\sqrt{3^{2} + 4^{2}}} = \frac{|6 - 12 - 5|}{5} = \frac{|-11|}{5} = \frac{11}{5}$.Since the axis is parallel to the $y$-axis,the equation of the parabola is $(x - h)^{2} = 4a(y - k)$,where $4a = \frac{11}{5}$.$(x - h)^{2} = \frac{11}{5}(y - k)$.Differentiating with respect to $x$:$2(x - h) = \frac{11}{5} \frac{dy}{dx}$.Differentiating again with respect to $x$:$2 = \frac{11}{5} \frac{d^{2}y}{dx^{2}}$.$10 = 11 \frac{d^{2}y}{dx^{2}}$,which is $11 \frac{d^{2}y}{dx^{2}} = 10$.

$A$ differential equation representing the family of parabolas with axis parallel to the $y$-axis and whose length of latus rectum is the distance of the point $(2, -3)$ from the line $3x + 4y = 5$,is given by:

Similar Questions

The differential equation obtained by eliminating the arbitrary constants $a$ and $b$ from $xy = ae^x + be^{-x}$ is

The differential equation of all circles which pass through the origin and whose centres lie on the $y$-axis is:

The differential equation of all circles having their centres on the line $y=5$ and touching the $X$-axis is ......

Verify that the given function $y = x^{2} + 2x + C$ is a solution of the differential equation $y' - 2x - 2 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module