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The equation of a curve is y^2+2x = 13 and the equation of a line is 2y+x = k, where k...

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saj-94

Posted September 3, 2013 at 7:57 AM via web

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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.

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llltkl | College Teacher

Posted September 3, 2013 at 8:19 AM (Answer #1)

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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)

`rArr y'=-1/2`

slope = -1/2

So we have to find where curve `y^2+2x=13` has slope = -1/2

`y^2+2x=13`

`rArr 2y*y'=-2`

`rArr y'=-1/y`

Put the condition for tangency,

`-1/2=-1/y`

`rArr y=2`

Put this value of y in the equation of the curve,

`2^2+2x=13`

`rArr x=9/2`

Now put the values of x and y to get k as:

`2*2+9/2=k`

`rArr k=17/2`

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` ).

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aruv | High School Teacher

Posted September 3, 2013 at 8:46 AM (Answer #2)

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Given equation of the curve is

`y^2+2x=13`

`` 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.

Thus

2y+x=K            (ii) ,

can intersect  (i)  at most two points. (i) intersect (ii) if

`((K-x)/2)^2+2x=13`  

`K^2+x^2-2Kx+8x=52`

`x^2+(8-2K)x+K^2-52=0`  (iii)

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.

`(8-2K)^2-4(K^2-52)=0`

`64+4K^2-32K-4K^2+208=0`

`-32K+272=0`

`-32K=-272`

`K=272/32`

`=8.5`

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.

 

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