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

Submitted by Bingo on July 8, 2017 - 5:29pm

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.

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## 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

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

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

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

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

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

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

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

I think you are from ADZU haha

## Good evening, sir. Do you

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)?