Question #a76f9

2 Answers
Feb 14, 2017

x+4y+-3=0. See the tangents-inclusive Socratic graph.

Explanation:

The equations have the form

y=-1/4x+c.

The tangent lines meet the hyperbola x y = 9 in two coincident

points.

Eliminating y and equating the discriminant of the quadratic

x(-1/4x+c)=9 to 0,

c^2-9=0, giving c = +-3.

graph{(xy-9)((y+x/4)^2-9)=0 [-40, 40, -20, 20]}
.

Feb 14, 2017

The equations of the two tangents that have gradient -1/4 are (without loss of generality):

l_1: \ \ \ y = -1/4x-3
l_2: \ \ \ y = -1/4x+3

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point.

The equation of the given hyperbola, H, is:

xy = 9

Differentiate (implicitly and product rule) the equation of the hyperbola:

(x)(dy/dx)+(1)(y) = 0
:. dy/dx = -y/x

A generic point on the hyperbola can be parametrised with a parameter t as follows:

x= t => yt=9 => y=9/t

So a generic point has coordinates P(t, 9/t). Thus, the gradient of the tangent at P is given by:

dy/dx = -y/x = -(9/t)/(t) =-9/t^2

At the point where the gradient of the tangent is -1/4 we have:

-9/t^2 = -1/4
:. t^2=36
:. t=+-6

(NB We expected two solutions as there are two tangents!)

Case 1
When t=-6 => P(t,1/t)=(-6,-3/2)
So, the tangent passes through (-6,-3/2) and has gradient -1/4,
so using the point/slope form of a straight line y-y_1=m(x-x_1) the equation we seek is;

y-(-3/2)=(-1/4)(x-(-6))
:. y+3/2 = -1/4x-3/2
:. y = -1/4x-3

Case 2
When t=6 => P(t,1/t)=(6,3/2)
So, the tangent passes through (6,3/2) and has gradient -1/4,
so using the point/slope form of a straight line y-y_1=m(x-x_1) the equation we seek is;

y-(3/2)=(-1/4)(x-6)
:. y-3/2 = -1/4x+3/2
:. y = -1/4x+3

Hence the equations of the two tangents that have gradient -1/4 are (without loss of generality):

l_1: \ \ \ y = -1/4x-3
l_2: \ \ \ y = -1/4x+3

We can verify this graphically:
enter image source here