# Paul's questions at Yahoo! Answers regarding tangent and normal lines

#### MarkFL

Staff member
Here are the questions:

Equation of tangent and normal?

1.find the equation of the tangent and normal to the curve y = x/a+a/x at (a/2, 5/2)

2. at what point of the parabola y=x^2-3x-5 is the tangent line parallel to 3x-y=2? find the eqn of the tangent line.

3. find the eqn. of the line normal to the curve y=3x^5+10x^3+15x+1 at its point of inflection.

4. find the critical points and the points of inflection of the curve y= 3x^4-8x^3+6x^2

Here is a link to the questions:

Equation of tangent and normal? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.

#### MarkFL

Staff member
Hello Paul,

1.) We are asked to find the equation of the tangent and normal lines to the curve $$\displaystyle y=\frac{x}{a}+\frac{a}{x}$$ at $$\displaystyle \left(\frac{a}{2},\frac{5}{2} \right)$$.

First, we need to compute the derivative of the curve to find the slopes:

$$\displaystyle \frac{dy}{dx}=\frac{1}{a}-\frac{a}{x^2}=\frac{x^2-a^2}{ax^2}$$

Hence, the slope of the tangent line is:

$$\displaystyle m=\left.\frac{dy}{dx} \right|_{x=\frac{a}{2}}=\frac{\left(\frac{a}{2} \right)^2-a^2}{a\left(\frac{a}{2} \right)^2}=-\frac{3}{a}$$

and so the slope of the normal line is:

$$\displaystyle -\frac{1}{m}=\frac{a}{3}$$

We now have the slopes and the point, so applying the point-slope formula, we find:

i) Tangent line:

$$\displaystyle y-\frac{5}{2}=-\frac{3}{a}\left(x-\frac{a}{2} \right)$$

$$\displaystyle y=-\frac{3}{a}x+4$$

ii) Normal line:

$$\displaystyle y-\frac{5}{2}=\frac{a}{3}\left(x-\frac{a}{2} \right)$$

$$\displaystyle y=\frac{a}{3}x+\frac{15-a^2}{6}$$

Here are some plots with a few values for $a$:

2.) We are asked to find at what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find the equation of the tangent line.

First we observe that we may arrange the given line in slope-intercept form as follows:

$$\displaystyle y=3x-2$$

Now, we want to equate the derivative of the parabola to the slope of the line (since parallel lines have equal slopes):

$$\displaystyle 2x-3=3$$

$$\displaystyle x=3$$

Now, to find the $y$-coordinate:

$$\displaystyle y(3)=(3)^2-3(3)-5=-5$$

and so we now have the slope $$\displaystyle m=3$$ and the point $$\displaystyle (3,-5)$$, and applying the point-slope formula, we find then that the equation of the tangent line parallel to the given line is:

$$\displaystyle y-(-5)=3(x-3)$$

$$\displaystyle y=3x-14$$

Here is a plot of the parabola, the given line, and the line tangent to the parabola and parallel to the given line:

3.) We are asked to find the equation of the line normal to the curve $y=3x^5+10x^3+15x+1$ at its point of inflection.

From the first derivative, we may determine the slope of the tangent line, once we know where the point of inflection is.

$$\displaystyle \frac{dy}{dx}=15x^4+30x^2+15=15\left(x^2+1 \right)^2$$

$$\displaystyle \frac{d^2y}{dx^2}=60x^3+60x=60x\left(x^2+1 \right)=0$$

We find that the second derivative has only the real root $x=0$ and since it is of multiplicity 1, we know the second derivative changes sign across this root, hence the point of inflection is at $$\displaystyle (0,1)$$. Now the slope of the normal line is:

$$\displaystyle m=-\frac{1}{15}$$ and so, applying the point slope formula, we find the normal line is:

$$\displaystyle y-1=-\frac{1}{15}(x-0)$$

$$\displaystyle y=-\frac{1}{15}x+1$$

4.) We are asked to find the critical points and the points of inflection of the curve $y=3x^4-8x^3+6x^2$.

Critical or stationary points:

We want to equate the first derivative to zero, and solve for $x$:

$$\displaystyle \frac{dy}{dx}=12x^3-24x^2+12x=12x(x-1)^2=0$$

The critical values are then:

$$\displaystyle x=0,\,1$$

Because the root $x=1$ is of multiplcity 2, we may conclude that this is not an extremum, and by the first derivative test we know the critical value $x=0$ is at a relative minimum. Thus $$\displaystyle (0,0)$$ is a relative minimum.

Points of inflection:

$$\displaystyle \frac{d^2y}{dx^2}=36x^2-48x+12=12\left(3x^2-4x+1 \right)=12(3x-1)(x-1)=0$$

Since the roots are of multiplcity 1, we know the sign of the second derivative changes across then, and so we may conclude that the points of inflection are at:

$$\displaystyle \left(\frac{1}{3},\frac{11}{27} \right)$$

$$\displaystyle (1,1)$$

Here is a plot of the quartic function, showing its minimum at the origin and the points of inflection:

To Paul and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.