Find the exact Value of c

Question

Find the exact Value of c on f(x) = c/x so that f(x) is tangent to x²/25 + y²/9 = 1

Answer 1

I will leave most of the algebra for you, but highlight the main steps.

 

Two functions f(x) and g(x) are tangent IF f(x)=g(x) AND f'(x) = g'(x);

That is the functions and their derivatives must agree.

 

To do this, the hyperbola y = f(x) = C/x gets plugged into the ellipse x^2/9 + y^2/25 = 1 where y=g(x)

 

f(x)=g(x)

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x^2/9 + (C/x)^2 / 25 = 1  <--- plugs y=C/x into the ellipse

 

x^2/9 + c^2/(25x^2) = 1

 

25x^4 + 9c^2 = 225x^2  <--- multiplies everything by 225x^2;

                                       <<--- label this equation ALPHA

 

f'(x)=g'(x)

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y' = f'(x) = -c*x^(-2) = -C/x^2  by power rule

 

Implicitly differentiating the ellipse:

   (2/9)x + (2/25)*y*y' = 0

 

Plugging in y= C/x and y' = -C/x^2,

 

(2/9)X + (2/25)(C/x)(-C/x^2) = 0

 

 

(2/9)X - (2/25C^2)(1/x^3) = 0

 

Multiplying everything by x^3:

 

(2/9)X^4 - (2/25C^2) = 0

 

So,

(2/9)X^4 = (2/25C^2) 

 

 

 

Multiplying both sides by 9/2:

 

X^4 = (9/25)C^2

 

X^2 = +or- (3/5)C

 

Plugging these into equation ALPHA:

 

25(9/25)C^2) + 9C^2 = 225(+or- (3/5)C )

 

9C^2 + 9C^2 =  +or- 135C

 

18C^2 +or- 135C = 0

 

9C ( 2C +or - 15) = 0

 

C = 0 or (2C +or- 15) = 0

 

C=0 does not produce any feasible solution as it is the minor axis of the ellipse.

 

So 2C +or- 15 = 0 --->  2C = +or- 15 ---> C = +or- 15/2 = +or- 7 and 1/2 = +or- 7.5

 

So the hyperola is y = +or- 15/(2X) 0r +or- 7.5/X

The tangent points are

    

[ 3*Sqrt(2)/2 , 5*sqrt(2)/2 ]   or APPROXIMATELY (2.121, 3.536)

 

and

 

[-3*sqrt(2)/2,  -5*sqrt(2)/2 ]  or APPROXIMATELY (-2.121, -3.536)