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Brandon's questions at Yahoo! Answers regarding tangent lines

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MarkFL

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Feb 24, 2012
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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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MarkFL

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Feb 24, 2012
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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:

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