Show that the line 4x - y + 11 = 0 is tangent to the curve y = 16/x^2 -1.  Any help please? I was thinking that you would find the derivative/slope of each equation and then equate them. By the...

Show that the line 4x - y + 11 = 0 is tangent to the curve y = 16/x^2 -1.

 

Any help please? I was thinking that you would find the derivative/slope of each equation and then equate them. By the way, my teacher didn't give any coordinate points.

Asked on by nerdynikki

1 Answer | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

It has to be shown that the line 4x - y + 11 = 0 is tangent to the curve y = 16/x^2 -1.

4x - y + 11 = 0

=> y = 4x + 11

Substituting in y = 16/x^2 -1

=> (4x + 11) = 16/x^2 - 1

=> (4x + 11)*x^2 = 16 - x^2

=> 4x^3 + 11x^2 = 16 - x^2

=> 4x^3 + 12x^2 - 16 = 0

=> x^3 + 3x^2 - 4  = 0

=> x^3 + 2x^2 + x^2 + 2x - 2x - 4 = 0

=> x^2(x + 2) + x(x + 2) - 2(x + 2) = 0

=> (x^2 + x - 2)(x + 2) = 0

=> (x^2 + 2x - x - 2)(x + 2) = 0

=> (x(x + 2) - 1(x + 2))(x + 2) = 0

=> (x - 1)(x + 2)^2 = 0

=> x = 1 and x = -2

The two curves intersect at the points where x = 1 and x = -2.

The derivative of y = 16/x^2 - 1 is -32/x^3

At x = -2, the value of the derivative is -32/-8 = 4

This is also the slope of the line 4x - y + 11 = 0

This shows that the line 4x - y + 11 = 0 is tangential to the curve y = 16/x^2 - 1 at a point.

We’ve answered 318,926 questions. We can answer yours, too.

Ask a question