# Find the differential equations of the following family of curves.

Differential Equations

Find the differential equations of the following family of curves.

1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.

2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.

3. All ellipses with center at the origin and axes on the coordinate axes.

4. Family of cardioids.

5. Family of 3 – leaf roses.

## For number 1.

1

## (1) Upwards and downwards

(1) Upwards and downwards parabolas with latus rectum equal to 4a.

$y - k = \pm 4a(x - h)^2$

$y' = \pm 8a(x - h)$

$y'' = \pm 8a$

## y−k=±4a(x−h)2 is this the

In reply to (1) Upwards and downwards by Jhun Vert

y−k=±4a

_{(x−h)}2 is this the equation for parabolas?isn't it 4a(y-k)=

_{(x-h)}2 ?## I made a mistake in there, it

In reply to y−k=±4a(x−h)2 is this the by Bingo

I made a mistake in there, it should be (

x-h)^{2}= ±4a(y-k). The solution for (1) should go this way:$(x - h)^2 = \pm 4a(y - k)$

$2(x - h) = \pm 4ay'$

$2 = \pm 4ay''$

$y'' = \pm \frac{1}{2a}$

## y−k=±4a(x−h)2 is this the

In reply to (1) Upwards and downwards by Jhun Vert

y−k=±4a

_{(x−h)}2 is this the equation for parabolas?isn't it 4a(y-k)=

_{(x-h)}2 ?## Yes you are right, please

In reply to y−k=±4a(x−h)2 is this the by Bingo

Yes you are right, please refer to my reply above

## so for number 2,

so for number 2,

_{x-h=±4a(y-k)}2x'=±8a(y-k)

x''=±8a

am i wrong?

## It is better to express your

In reply to so for number 2, by Bingo

It is better to express your answer in terms of

y'rather thanx'. Althoughx'will do and simpler.$(y − k)^2 = \pm 4a(x − h)$

$2(y − k)y' = \pm 4a$

$y' = \dfrac{\pm 2a}{y - k}$

$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$

$y'' = \dfrac{\mp 8a}{y'}$

$y'' \, y' = \pm 8a$

## our prof. gave the same

In reply to so for number 2, by Bingo

our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?

## In the above solution, I made

In reply to our prof. gave the same by ChaCha Ingal Cortez

They are different. I made a mistake in $y - k$ of the above solution.

$(y − k)^2 = \pm 4a(x − h)$

$2(y - k) \, y' = \pm 4a$

$y' = \dfrac{\pm 2a}{y - k}$

$y - k = \dfrac{\pm 2a}{y'}$

$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$

$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$

$y'' = \dfrac{\mp (y')^3}{2a}$

$2a \, y'' = \mp (y')^3$

$\pm (y')^3 + 2a \, y'' = 0$

## Thank You so much sir.

Thank You so much sir.

## sir, do you know the standard

sir, do you know the standard or general equations for the last 3 problems?

what equations should i use?

## I think you are from ADZU

In reply to sir, do you know the standard by Bingo

I think you are from ADZU haha

## Good evening, sir. Do you

In reply to I think you are from ADZU by Shandy

Good evening, sir. Do you have the solution for the last three problems? Thank you!

## Sir do you know how to

Sir do you know how to eliminate arbitrary constant here in this equation?

(y-33)²=4a(x-h)?