Please Help !

Question

there is given the hyperbole x²/49-y²/16=1. Find the point M which distance from one asymptote is three times longer than from the other asymptote of the hyperbole.

Answer 1

This is a somewhat ugly problem. I'm not sure what math class you're in, but I hope you know the formulas involved.

First....the asymptotes of your hyperbola has the equations y=4/7 x and y = -4/7 x. Written in standard form this is

4x + 7y = 0 and -4x + 7y = 0. We need it in this form so that we can use the formula for the distance a point is from a line. That formula is as follows:

 

The distance point (m,n) is from line Ax + By + C = 0 is Absolute value of Am+Bn +C divided by Sqrt(A^2 + B^2).

So our 2 distances are ABS(4m + 7n)/sqrt(65) and ABS(-4m + 7n)/Sqrt(65).

 

Now one distance is 3 times the other leads to the equation

ABS(4m+7n)/sqrt(65) = 3*ABS(-4m+7n)/sqrt(65)     this can be broken into 2 equations

 

(4m+7n)/sqrt65 = 3(-4m+7n)/sqrt65                  or        (4m+7n)/sqrt65 = 3(4m-7n)/sqrt65

 

The first equation leads to imaginary answers.

The second equation leads to n = (-2/7)m. This must satisfy your hyperbola equation so that the point is on the hyperbola.

Substituting we get.....note that m is the x coordinate and n is the y coordinate,

 

m^2/49 - (4/49)m^2/16 =1   solving this we get

4m^2/196 - m^2/196 =1

3m^2/196 = 1

m^2 = 196/3

m = + 14/sqrt3 or - 14/sqrt3

 

Using the positive value of m, This makes n = (-2/7)(+14/sqrt3) = -4/sqrt3

So one possible coordinate is ( 14/sqrt3 , -4/sqrt3).

 

I checked these and they are on the hyperbola and one distance IS 3 times the other.

 

Hope this helped.