Brandon's questions at Yahoo Answers regarding tangent lines

In summary, in the first question, we are given a curve and a point on the curve and are asked to find the value of a constant that would make the tangent line at that point pass through another given point. In the second question, we are asked to find the equation of the tangent line at a specific point, determine the values of a for which this tangent line passes through a given point, and find the possible equations of the tangent line from that given point to the curve.
  • #1
MarkFL
Gold Member
MHB
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Here are the questions:

Tangents, Normal. Need some help.?


I just recently took a test, and i am unsure of the following questions. It would be great if you provide me some explanations and workings.

1) A curve has the equation y = 2x^(2) - kx + 3, where k is a constant. The tangent at the point A passes through the point B(5,1). Find the value of k.

2) A curve has the equation y = x + x^(2). Find

i) The equation of the tangent to the curve at the point where x = a.

ii) The value(s) of a for which this tangent passes through the point P(-2,3)

iii) Hence find the possible equations of the tangent from P to the curve.

Thank you so much guys! Have a great day

I have posted a link there to this thread so the OP can view my work.
 
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  • #2
Hello Brandon,

1.) We are given the curve:

\(\displaystyle y=2x^2-kx+3\) where $k$ is a constant.

Let point $A$ be given by:

\(\displaystyle \left(x_A,y_A \right)=\left(x_A,2x_A^2-kx_A+3 \right)\)

The slope of the line passing through the points $A$ and $B$ is:

\(\displaystyle m=\frac{\left(2x_A^2-kx_A+3 \right)-(1)}{\left(x_A \right)-(5)}=\frac{2x_A^2-kx_A+2}{x_A-5}\)

Also, given that:

\(\displaystyle m=\left.\frac{dy}{dx} \right|_{x=x_A}=4x_A-k\)

We may then state:

\(\displaystyle \frac{2x_A^2-kx_A+2}{x_A-5}=4x_A-k\)

Multiplying through by \(\displaystyle x_A-5\), we obtain:

\(\displaystyle 2x_A^2-kx_A+2=4x_A^2-(20+k)x_A+5k\)

Combining like terms, we may arrange this as:

\(\displaystyle 5k=2+20x_A-2x_A^2\)

Hence:

\(\displaystyle k=\frac{2}{5}\left(1+10x_A-x_A^2 \right)\)

2.) We are given the curve:

\(\displaystyle y=x+x^2\)

i) First we find the slope of the tangent line at $x=a$ is:

\(\displaystyle m=\left.\frac{dy}{dx} \right|_{x=a}=2a+1\)

And this tangent line must pass through the point:

\(\displaystyle \left(a,a+a^2 \right)\)

Thus, using the point-slope formula, we find the equation of the tangent line is given by:

\(\displaystyle y-\left(a+a^2 \right)=(2a+1)(x-a)\)

Distributing on the right side, we get:

\(\displaystyle y-\left(a+a^2 \right)=(2a+1)x-2a^2-a\)

Adding $a+a^2$ to both sides, we get the tangent line in slope-intercept form:

\(\displaystyle y=(2a+1)x-a^2\)

ii) If this tangent line passes through the point $(-2,3)$, then we must have:

\(\displaystyle 3=(2a+1)(-2)-a^2\)

Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:

\(\displaystyle a^2+4a+5=0\)

We see the discriminant is negative, and so we conclude there are no real values of $a$ for which a tangent line to the given quadratic curve will pass through point $B$.

iii) There are no such possible tangent lines as we found in part ii).

I suspect that point $P$ was incorrectly given. I will consider the two following cases:

a) Point $P$ is supposed to be $(-2,-3)$. Then part ii) becomes:

\(\displaystyle -3=(2a+1)(-2)-a^2\)

Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:

\(\displaystyle a^2+4a-1=0\)

Application of the quadratic formula yields:

\(\displaystyle a=-2\pm\sqrt{5}\)

And so the two tangent lines would be given by:

\(\displaystyle y=\left(2\left(-2\pm\sqrt{5} \right)+1 \right)x-\left(-2\pm\sqrt{5} \right)^2=\left(-3\pm2\sqrt{5} \right)x-\left(9\pm4\sqrt{5} \right)\)

b) Point $P$ is supposed to be $(2,3)$. Then part ii) becomes:

\(\displaystyle 3=(2a+1)(2)-a^2\)

Now we may solve for $a$. Arranging the quadratic in $a$ in standard form, we obtain:

\(\displaystyle a^2-4a+1=0\)

Application of the quadratic formula yields:

\(\displaystyle a=2\pm\sqrt{3}\)

And so the two tangent lines would be given by:

\(\displaystyle y=\left(2\left(2\pm\sqrt{3} \right)+1 \right)x-\left(2\pm\sqrt{3} \right)^2=\left(5\pm2\sqrt{3} \right)x-\left(7\pm4\sqrt{3} \right)\)
 

FAQ: Brandon's questions at Yahoo Answers regarding tangent lines

1. What are tangent lines?

A tangent line is a straight line that touches a curve at only one point, without crossing through it. It represents the instantaneous rate of change of the curve at that point.

2. How do you find the equation of a tangent line?

To find the equation of a tangent line, you need to know the slope of the line and a point that lies on the line. The slope can be found using the derivative of the curve, and the point can be found by plugging in the x-coordinate of the point of tangency into the original equation.

3. Can a curve have more than one tangent line at a given point?

Yes, a curve can have more than one tangent line at a given point if the curve has a sharp turn or a cusp at that point. In these cases, the curve may have two or more tangent lines with different slopes.

4. How are tangent lines used in real-world applications?

Tangent lines are used in many real-world applications, such as physics, engineering, and economics. In physics, they are used to represent the velocity of an object at a given point in time. In engineering, they are used to find the slope of a curve on a graph, which can help in designing structures. In economics, they are used to represent the marginal cost or revenue of a product at a specific quantity.

5. What is the difference between a secant line and a tangent line?

A secant line is a line that intersects a curve at two or more points, while a tangent line only touches the curve at one point. A secant line can be used to find the average rate of change of a curve over an interval, while a tangent line represents the instantaneous rate of change at a single point.

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