The equation of a curve is y^2+2x = 13 and the equation of a line is 2y+x = k, where k is a constant. (ii) Find the value of k for which the line is a tangent to the curve.
Given equation of the curve is
`` which can be rewritten as
`y^2=-2x+13` (i) , which is an equation of parabola.
A straight line intersect parabola at most two points.
2y+x=K (ii) ,
can intersect (i) at most two points. (i) intersect (ii) if
has real roots i.e x has real values. Since (ii) is tangents to (i),therefore roots of the equation (ii) are equal.
roots will equal if discriminant of quadratic equation (iii) in x is zero.
Thus K=8.5 then line 2y+x=k 2 will tangent to the given curve.
Black line K>8.5
green line k=8.5
blue line k<8.5
See green is tangent to the curve.
First, the slope of the line `2y+x = k` has to be found.
`2y+x = k`
`rArr y=1/2(k-x)` (where k is a constant)
slope = -1/2
So we have to find where curve `y^2+2x=13` has slope = -1/2
Put the condition for tangency,
Put this value of y in the equation of the curve,
Now put the values of x and y to get k as:
Therefore, for `k=17/2` , the line `2y+x=k` is tangent to the curve `y^2+2x=13` , at the point (`9/2, 2` ).