The point $P (α, β)$ $P(alpha,beta)$ lies on the curve $C$ $C$ with equation $f (x) = \frac{1}{3} x + \frac{12}{x}$ $f(x)= 1/3x+12/x$ such that $α > 0, α \neq 6$ $alpha>0, alpha≠6$ . The normal at $P$ $P$ only crosses $C$ $C$ once. How do I find the exact value of $α$ $alpha$ ?

Question

The point $P (α, β)$ $P(alpha,beta)$ lies on the curve $C$ $C$ with equation $f (x) = \frac{1}{3} x + \frac{12}{x}$ $f(x)= 1/3x+12/x$ such that $α > 0, α \neq 6$ $alpha>0, alpha≠6$ . The normal at $P$ $P$ only crosses $C$ $C$ once. How do I find the exact value of $α$ $alpha$ ?

3 Answers

Impact of this question

1479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-05T18:57:01

There is no such point

Explanation:

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 normal is perpendicular to the tangent so the product of their gradients is $- 1$ $-1$

We have:

$f(x) = x/3+12/x \ \ \ \ ..... (star)$

$" " = (x^2+36)/(3x)$

First we find the coordinate $beta$ ; At $P(alpha,beta)$ we have:

$beta = f(alpha)$
$\ \ = (alpha^2+36)/(3alpha)$

Then differentiating $(star)$ wrt $x$ gives us:

$f'(x) = 1/3-12/x^2$

$" " = (x^2-36)/(3x^2)$

So at the point $P(alpha,beta)$ the gradient of the tangent is given by:

$m_T = f'(alpha)$
$" " = (alpha^2-36)/(3alpha^2)$

Hence, the gradient of the normal at $P(alpha,beta)$ is given by:

$m_N = -1/m_T$
$" " = -1/((alpha^2-36)/(3alpha^2))$
$" " = -(3alpha^2)/(alpha^2-36) \ \ \ \ because alpha != 6$

So the normal passes through $P(alpha,beta)$ and has gradient $m_N$ , so using the point/slope form $y-y_1=m(x-x_1)$ the equation of the normal at $P(alpha,beta)$ is:

$y-beta = -(3alpha^2)/(alpha^2-36) (x - alpha)$

$:. y= (3alpha^2(alpha-x))/(alpha^2-36) + (alpha^2+36)/(3alpha)$

$:. y= (9alpha^3(alpha-x) + (alpha^2+36)(alpha^2-36))/((3alpha)(alpha^2-36))$

$:. y= (9alpha^4-9alpha^3x + alpha^4-1296)/((3alpha)(alpha^2-36))$

$:. y = (10alpha^4-9alpha^3x -1296)/((3alpha)(alpha^2-36))$

This normal at $P$ will cross the original curve $y=f(x)$ when we have a simultaneous solution of both of the equations:

${ (y = (x^2+36)/(3x), "(The Function)" ), ( y = (10alpha^4-9alpha^3x -1296)/((3alpha)(alpha^2-36)), "(The Normal)" ) :}$

i.e. a solutions of:

$(x^2+36)/(3x) = (10alpha^4-9alpha^3x -1296)/((3alpha)(alpha^2-36))$

$:. (x^2+36)/(x) = (10alpha^4-9alpha^3x -1296)/(alpha^3-36alpha)$

$:. (x^2+36)(alpha^3-36alpha) = (10alpha^4-9alpha^3x -1296)x$

$:. alpha^3x^2-36alphax^2+36alpha^3-1296alpha = 10alpha^4x-9alpha^3x^2 -1296x$

$:. alpha^3x^2-36alphax^2+36alpha^3-1296alpha - 10alpha^4x+9alpha^3x^2 +1296x =0$

$:. (10alpha^3-36alpha)x^2 + (1296- 10alpha^4)x+(36alpha^3-1296alpha) = 0$

$:. (5alpha^3-18alpha)x^2 + (648- 5alpha^4)x+(18alpha^3-648alpha) = 0$

This is a quadratic in $x$ , and so either has:

one repeated real solution

two unique real solutions

two complex solutions

We want our Normal to touch the original curve just once, and therefore we require one repeated real solution, and so the discriminant, $Delta=b^2-4ac$ of the quadratic must be zero, this requires:

$(648- 5alpha^4)^2-4(5alpha^3-18alpha)(18alpha^3-648alpha) = 0$

$:. 419904-6480alpha^4+25alpha^8 - 4(90alpha^6-3240alpha^4-324alpha^4+11664alpha^2) = 0$

$:. 419904-6480alpha^4+25alpha^8 -360alpha^6+14256alpha^4-46656alpha^2 = 0$

$:. 25alpha^8-360alpha^6+7776alpha^4-46656alpha^2 +419904= 0$

$:. (5alpha^4-36alpha^2+648)^2= 0$

$:. 5alpha^4-36alpha^2+648= 0$

And for this quadratic equation in $alpha^2$ , we note that its discriminant is:

$Delta = (-36)^2-4(5)(648)$
$\ \ \ = 1296-12960$
$\ \ \ = -11664$
$\ \ \ < 0$

There are no real solutions for $alpha^2$ , and therefore no real solutions for $alpha$

Thus the premise of the original question is false, and there is no such point $P(alpha,beta)$ that satisfies the given condition.

Answer 2 · 2017-05-06T00:06:13

$alpha = 3 sqrt[2/5]$

Explanation:

The function

$f(x) = x/3+12/x$ has two leafs and two asymptotes . One vertical at $x = 0$ and other slanted which is

$y=x/3$

Building a normal line in the positive-right leaf, we have a point which is it's minimum point, such that the normal does intersect $f(x)$ only once. This point is at

$(df)/(dx)=1/3 - 12/x_0^2=0$ giving $x_0 = pm 6$

The only non intersecting normal for $x ne 6$ is given at

$-1/((df)/(dx)) = 1/3$

or

$(df)/(dx) = -3$

or

$1/3 +3- 12/x_1^2 = 0$

so for $x_1 =pm3 sqrt[2/5]$

The normal equations are then

$y_1=sqrt[2/5] + 2 sqrt[10] + 1/3 (x-3 sqrt[2/5] )$
$y_2=-sqrt[2/5] - 2 sqrt[10] + 1/3 (x+3 sqrt[2/5] )$

So, we can construct four normal lines which do not intersect the other leaf. The verticals and the slanted.

Attached a plot showing the normal lines at $x = x_1 = pm3 sqrt[2/5]$

enter image source here

Answer 3 · 2017-05-06T15:44:17

See below.

Explanation:

Another approach.

Defining $p = (x,y)$ and $p_0=(alpha,beta)$

$f(x,y)=f(p) = y-x/3-12/x =0$

we have

$vec n_0 = grad f(p_0) = (-1/3 + 12/alpha^2, 1)$

so the normal line passing by $p_0$ is

$L_n->p=p_0+lambda vec n_0$ or

${(alpha -lambda/3 + (12 lambda)/alpha^2=0),(12/alpha + alpha/3 + lambda=0):}$

Now doing

$f(p_0+lambda vec n_0)=0$ and solving for $lambda$ we have

${(lambda = 0), (lambda= ( 3 alpha (648 - 36 alpha^2 + 5 alpha^4))/( 648 - 198 alpha^2 + 5 alpha^4)):}$

So we have the two potential intersections within the normal and the function $f(p)=0$

One of them is obviously for $lambda = 0$ . If this is the only intersection, then

$lambda= ( 3 alpha (648 - 36 alpha^2 + 5 alpha^4))/( 648 - 198 alpha^2 + 5 alpha^4)$

must be not a real number. This is satisfied when

$648 - 198 alpha^2 + 5 alpha^4=0$ with the solutions

$alpha = (-6,6,-3 sqrt[2/5],3 sqrt[2/5])$

so there are solutions located at $alpha = pm 3 sqrt[2/5]$

Science

Math

Humanities

... and beyond

The point $P (α, β)$ $P(alpha,beta)$ lies on the curve $C$ $C$ with equation $f (x) = \frac{1}{3} x + \frac{12}{x}$ $f(x)= 1/3x+12/x$ such that $α > 0, α \neq 6$ $alpha>0, alpha≠6$ . The normal at $P$ $P$ only crosses $C$ $C$ once. How do I find the exact value of $α$ $alpha$ ?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

The point P(alpha,beta)P(α,β)P(alpha,beta) lies on the curve CCC with equation f(x)= 1/3x+12/xf(x)=13x+12xf(x)= 1/3x+12/x such that alpha>0, alpha≠6α>0,α≠6alpha>0, alpha≠6. The normal at PPP only crosses CCC once. How do I find the exact value of alphaαalpha?

3 Answers

Explanation:

Explanation:

Explanation:

Related questions

Impact of this question

The point $P (α, β)$ $P(alpha,beta)$ lies on the curve $C$ $C$ with equation $f (x) = \frac{1}{3} x + \frac{12}{x}$ $f(x)= 1/3x+12/x$ such that $α > 0, α \neq 6$ $alpha>0, alpha≠6$ . The normal at $P$ $P$ only crosses $C$ $C$ once. How do I find the exact value of $α$ $alpha$ ?