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distance between parabola endpoints

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3

$begingroup$

I've seen a question asked in an interview as following.

How can be the distance indicated by question mark calculated?
What are the ways?

geometry graphing-functions

share|cite|improve this question

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does the 80m stand for? Is it the length of the parabola?
  $endgroup$
  – Chubby Chef
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChubbyChef yes.
  $endgroup$
  – itsnotmyrealname
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a question in a rather similar spirit: How far apart are the towers?
  $endgroup$
  – Martin Sleziak
  1 hour ago
  

  
  
  
  

add a comment | 

3

$begingroup$

I've seen a question asked in an interview as following.

How can be the distance indicated by question mark calculated?
What are the ways?

geometry graphing-functions

share|cite|improve this question

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What does the 80m stand for? Is it the length of the parabola?
  $endgroup$
  – Chubby Chef
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @ChubbyChef yes.
  $endgroup$
  – itsnotmyrealname
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Here is a question in a rather similar spirit: How far apart are the towers?
  $endgroup$
  – Martin Sleziak
  1 hour ago
  

  
  
  
  

add a comment | 

$begingroup$

I've seen a question asked in an interview as following.

How can be the distance indicated by question mark calculated?
What are the ways?

geometry graphing-functions

share|cite|improve this question

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

$endgroup$

share|cite|improve this question

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

share|cite|improve this question

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

asked 8 hours ago

itsnotmyrealnameitsnotmyrealname

18110

4 Answers
4

active

oldest

votes

5

$begingroup$

The distance would be $0$ meters.

The ends of the parabola are $50$ meters high and the parabola itself $80$ meters in length. If you were to hold both ends of the parabola from the same point at that height, it would fall $80/2 = 40$ meters down, $10$ meters above the ground.

This is less any sort of involved detailed computation than it is a way to see how cleverly you can think.

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. "This is less any sort of involved detailed computation than it is a way to see how cleverly you can think." Indeed, the problem does not even indicate that the "curve" is actually a parabola; it could be a catenary (which would be appropriate if we're considering a rope or cable hanging across a chasm), or an arch of a sinusoid, or, or, or ... So, either the question is egregiously ill-posed (which is not unheard-of in interview questions or social media challenges) or it's being a little sneaky. Here, it's the latter.
  $endgroup$
  – Blue
  3 hours ago
  
  

  
  
  
  

add a comment | 

3

$begingroup$

Distance is $0$. The midpoint is at $40$ from the end. The midpoint height is $10$, so you need at least $40$ to get from top to middle, if the distance is $0$. If the distance is larger, then you need more cable.

share|cite|improve this answer

answered 8 hours ago

AndreiAndrei

14.9k21430

$endgroup$

add a comment | 

3

$begingroup$

hint

the equation of the parabola will be of the form
$$y=ax^2+10$$
where $-ble xle b,$
$$ab^2+10=50$$

and

$$L=80=2int_0^bsqrt1+4a^2x^2dx$$
the length you look for is

$$l=2b$$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

$endgroup$

add a comment | 

0

$begingroup$

I see the video mentioned in the comment but wonder if this would be of any help-

I fixed the coordinate frame marked in orange - green axis as the $x-y$ axis. Now I can get the coordinates of the points $A$ and $B$ marked in the figure below. Next, as parabolas are symmetric at the point $A$ with respect to $Y$ axis. So the horizontal distance between $A$ and $B$ is $50 -0$ and the required distance would be double of it which is 100.

share|cite|improve this answer

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

$endgroup$

add a comment | 

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5

$begingroup$

The distance would be $0$ meters.

The ends of the parabola are $50$ meters high and the parabola itself $80$ meters in length. If you were to hold both ends of the parabola from the same point at that height, it would fall $80/2 = 40$ meters down, $10$ meters above the ground.

This is less any sort of involved detailed computation than it is a way to see how cleverly you can think.

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. "This is less any sort of involved detailed computation than it is a way to see how cleverly you can think." Indeed, the problem does not even indicate that the "curve" is actually a parabola; it could be a catenary (which would be appropriate if we're considering a rope or cable hanging across a chasm), or an arch of a sinusoid, or, or, or ... So, either the question is egregiously ill-posed (which is not unheard-of in interview questions or social media challenges) or it's being a little sneaky. Here, it's the latter.
  $endgroup$
  – Blue
  3 hours ago
  
  

  
  
  
  

add a comment | 

5

$begingroup$

The distance would be $0$ meters.

The ends of the parabola are $50$ meters high and the parabola itself $80$ meters in length. If you were to hold both ends of the parabola from the same point at that height, it would fall $80/2 = 40$ meters down, $10$ meters above the ground.

This is less any sort of involved detailed computation than it is a way to see how cleverly you can think.

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  +1. "This is less any sort of involved detailed computation than it is a way to see how cleverly you can think." Indeed, the problem does not even indicate that the "curve" is actually a parabola; it could be a catenary (which would be appropriate if we're considering a rope or cable hanging across a chasm), or an arch of a sinusoid, or, or, or ... So, either the question is egregiously ill-posed (which is not unheard-of in interview questions or social media challenges) or it's being a little sneaky. Here, it's the latter.
  $endgroup$
  – Blue
  3 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

The distance would be $0$ meters.

The ends of the parabola are $50$ meters high and the parabola itself $80$ meters in length. If you were to hold both ends of the parabola from the same point at that height, it would fall $80/2 = 40$ meters down, $10$ meters above the ground.

This is less any sort of involved detailed computation than it is a way to see how cleverly you can think.

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

answered 8 hours ago

Eevee TrainerEevee Trainer

13k32046

3

$begingroup$

Distance is $0$. The midpoint is at $40$ from the end. The midpoint height is $10$, so you need at least $40$ to get from top to middle, if the distance is $0$. If the distance is larger, then you need more cable.

share|cite|improve this answer

answered 8 hours ago

AndreiAndrei

14.9k21430

$endgroup$

add a comment | 

3

$begingroup$

Distance is $0$. The midpoint is at $40$ from the end. The midpoint height is $10$, so you need at least $40$ to get from top to middle, if the distance is $0$. If the distance is larger, then you need more cable.

share|cite|improve this answer

answered 8 hours ago

AndreiAndrei

14.9k21430

$endgroup$

add a comment | 

$begingroup$

Distance is $0$. The midpoint is at $40$ from the end. The midpoint height is $10$, so you need at least $40$ to get from top to middle, if the distance is $0$. If the distance is larger, then you need more cable.

share|cite|improve this answer

answered 8 hours ago

AndreiAndrei

14.9k21430

$endgroup$

share|cite|improve this answer

answered 8 hours ago

AndreiAndrei

14.9k21430

answered 8 hours ago

AndreiAndrei

14.9k21430

answered 8 hours ago

AndreiAndrei

14.9k21430

3

$begingroup$

hint

the equation of the parabola will be of the form
$$y=ax^2+10$$
where $-ble xle b,$
$$ab^2+10=50$$

and

$$L=80=2int_0^bsqrt1+4a^2x^2dx$$
the length you look for is

$$l=2b$$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

$endgroup$

add a comment | 

3

$begingroup$

hint

the equation of the parabola will be of the form
$$y=ax^2+10$$
where $-ble xle b,$
$$ab^2+10=50$$

and

$$L=80=2int_0^bsqrt1+4a^2x^2dx$$
the length you look for is

$$l=2b$$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

$endgroup$

add a comment | 

$begingroup$

hint

the equation of the parabola will be of the form
$$y=ax^2+10$$
where $-ble xle b,$
$$ab^2+10=50$$

and

$$L=80=2int_0^bsqrt1+4a^2x^2dx$$
the length you look for is

$$l=2b$$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

$endgroup$

share|cite|improve this answer

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

answered 8 hours ago

hamam_Abdallahhamam_Abdallah

38.4k21634

0

$begingroup$

I see the video mentioned in the comment but wonder if this would be of any help-

I fixed the coordinate frame marked in orange - green axis as the $x-y$ axis. Now I can get the coordinates of the points $A$ and $B$ marked in the figure below. Next, as parabolas are symmetric at the point $A$ with respect to $Y$ axis. So the horizontal distance between $A$ and $B$ is $50 -0$ and the required distance would be double of it which is 100.

share|cite|improve this answer

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

$endgroup$

add a comment | 

0

$begingroup$

I see the video mentioned in the comment but wonder if this would be of any help-

I fixed the coordinate frame marked in orange - green axis as the $x-y$ axis. Now I can get the coordinates of the points $A$ and $B$ marked in the figure below. Next, as parabolas are symmetric at the point $A$ with respect to $Y$ axis. So the horizontal distance between $A$ and $B$ is $50 -0$ and the required distance would be double of it which is 100.

share|cite|improve this answer

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

$endgroup$

add a comment | 

$begingroup$

I see the video mentioned in the comment but wonder if this would be of any help-

I fixed the coordinate frame marked in orange - green axis as the $x-y$ axis. Now I can get the coordinates of the points $A$ and $B$ marked in the figure below. Next, as parabolas are symmetric at the point $A$ with respect to $Y$ axis. So the horizontal distance between $A$ and $B$ is $50 -0$ and the required distance would be double of it which is 100.

share|cite|improve this answer

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

$endgroup$

share|cite|improve this answer

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

answered 8 hours ago

BAYMAXBAYMAX

3,18021327

answered 8 hours ago

BAYMAXBAYMAX

3,18021327