Question 824635
Let point P be (x, y).
Then the distance from P to (4, 0), using the distance formula, is:
{{{sqrt((x-4)^2+(y-0)^2)}}} or just {{{sqrt((x-4)^2+y^2)}}}
Since...<ul><li>x = 1 is a vertical line; and</li><li>since distance from a point to a line is measured perpendicularly; and</li><li>since perpendicular to vertical is horizontal</li></ul>.. the distance from point P to the line x = 1 is a horizontal distance. And horizontal distances are calculated by subtracting the x-coordinates.
So the displace from point P to the line x = 1 is: {{{abs(x-1)}}}
(Note: Absolute value is used because distance should be positive. And without the absolute value x-1 would be negative if x < 1.)<br>
We are now ready to translate "A point P moves such that it is always twice as far from the points (4,0) as it is on the line x = 1" into an equation. The translation might be easier to see if we reword the quoted sentence to: "The distance from point P to the point (4, 0) is twice as much as the distance from point P to the line x = 1."
{{{sqrt((x-4)^2+y^2) = 2*abs(x-1)}}}
Now we simplify. Squaring the left side should be simple. And squaring the right side is not hard, either, since x-1, after we square it, could not be negative. So the absolute value is no longer needed!
{{{(sqrt((x-4)^2+y^2))^2 = (2*abs(x-1))^2}}}
{{{(x-4)^2+y^2 = 2^2*(abs(x-1))^2}}}
{{{x^2-8x+16+y^2 = 4*(x-1)^2}}}
{{{x^2-8x+16+y^2 = 4*(x^2-2x+1)}}}
{{{x^2-8x+16+y^2 = 4x^2-8x+4}}}
Now we put the equation in standard form (for conic sections). Subtracting the entire left side from both sides we get:
{{{0 = 3x^2-y^2-12}}}
This is the equation for part a.<br>
Looking at our equation, with no xy terms and with the x and y squared terms having opposite signs, we should recognize that it is the equation for a hyperbola.<br>
To graph this hyperbola it will help to put it into the {{{(x-h)^2/a^2-(y-k)^2/b^2 = 1}}} form for hyperbolas. Adding 12 to each side we get:
{{{12 = 3x^2-y^2}}}
Dividing both sides by 16 (in order to turn it into a 1):
{{{1 = 3x^2/12-y^2/12}}}
Reducing the the first fraction:
{{{1 = x^2/4-y^2/16}}}
Rewriting the numerators to fit the form:
{{{1 = (x-0)^2/4-(y-0)^2/16}}}<br>
We can now "read" the key data:<ul><li>With the x squared in front of the subtraction, this hyperbola has a horizontal transverse axis.</li><li>The "h" and "k" are zeros. So the center of the hyperbola is (0, 0) (aka the origin).</li><li>{{{a^2 = 4}}}. This makes {{{a = 2}}} Since "a" is the distance from the center to the vertices on the transverse axis and since center is (0, 0), the vertices are (2, 0) and (-2, 0).</li><li>{{{b^2 = 16}}} so b = 4.</li><li>With a = 2 and b = 4, the slopes of the asymptotes are: <u>+</u>{{{4/2}}} = <u>+</u>2</li></ul>I'll leave the graphing up to you. Here's what it should look like:
{{{graph(400, 400, -8, 8, -8, 8, sqrt(3x^2-12), -sqrt(3x^2-12))}}}
(Don't mind the different colors or the gaps (which are there only because algebra.com's graphing software doesn't handle nearly vertical parts of a graph well.)