How do I determine C, so that the straight line joining (0,3) and (5,2) is tangent to the curve y = c/(x+1) ?
The value of c is required such that the line joining the points (0,3) and (5,2) is tangent to the curve y = c/(x+1).
The equation of the line joining (0,3) and (5,2) is given by (y - 3)/(x - 0) = (2-3)/(5 - 0)
or (y - 3)*5 = -1*x
or x = 15 - 5y
Now this line is a tangent to y = c/(x +1)
y = c/(x+1)
y*(x + 1) = c
Replace x by 15 - 5y
y*(15 - 5y + 1) = c
16y - 5y^2 - c = 0
As the line is tangent to the curve, 16y - 5y^2 - c = 0 has only one solution. An equation ax^2 + bx + c = 0 has one root when b^2 = 4ac. For the equation derived this is the case when 16^2 - (4*-5*-c) = 0
c = 256/20 = 12.8
The required value of c is 12.8
Given curve is y= c/(x+1)
You need to find the equation of the straight line and it cutsthese two points, so the general formula for linear equation is y=mx+c where m is the slope and c is the y-intercept.
First, you need to sub. these two points into the equation
So the equation is y=-1/5x+3 - eqn1
At the tangent line, the slope of the line and the quadratic curve are similar so:
5c= (x+1)^2 -eqn2
Also, the curve and line intersect each other, so they equal each other. Using simultaneous equations:
5c= -x^2+14x+15 -eqn 3
Since eqn 3 and eqn 2 are equal so:
x=7or-1 (rejected because negative number)
To find C, sub. x=7 to eqn 3
Given curve y=c/(x+1).. (1)
Its slope at any poit is dy/dx = -c/(x+1)^2 (2)
The slope and equation of the line joining (0,3) and (5,2) are
(2-3)/(5)= -(1/5) ....(3) and
y = -(1/5)x+3........(4).
At the tangent point the slope of line and the curve are same, therefore, from(3) and (2): -1/5=-c/(x+1)^2 ==>
Solving for the line and the curve, from rom ((4) and (1),
-(1/5)x +3 = c/(x+1)==>
5c=(-x+15)(x+1) = -x^2+14x+15.........(7)
Eliminatin c between (6) and (7) we get x cordinate of the tangent touching the curve :
x=-1 and x=7. But x=-1 is a vertical asymptote.
Putting x=7 in Eq (6) gives 5c =(7+1)^2
c = 64/5 =12.8.
The tangent touches curve at
the point. (7, 12/(7+1))=(7, 1.5)
First of all, you have to note that the straight line is tangent to the curve, but only if the system formed by the equation of the straight line and the curve is unique determined (i.e., it has only one solution).
So, we have to determine the equation of the straight lline.
For that, we know that the straight line has on it the pair of coordinates: (0,3) and(5,2). We will write the general form of a linear equation (function).
y=mx + n
3=m*0 + n ; n=3
2=5m + n; 2= 5m + 3;
y=(-1/5)x + 3
But y = c/(x+1)
Now we substitute y = c/(x+1) in the equation of the straight line:
c/(x+1)= (-1/5)x + 3
The common denominator is 5(x+1)
5c= -x^2 + 14x + 15
x^2 - 14x - 15 + 5c = 0
In order to obtain a tangent point, the values of x have to be the same, so the discriminant delta is:
delta= (-14)^2 -4*1*(5c-15)=0
Opening the parantheses: