Question 823774
Let (x, y) be the point P. Then the distance from point P to (4, 0) would be (using the distance formula):
{{{sqrt((x-4)^2+(y-0)^2)}}}
which simplifies to:
{{{sqrt(x^2-8x+16+y^2)}}}<br>
The distance between a point and a line is always measured perpendicularly. Since the given line x = 1 is a vertical line, we will measure the distance from point P to the line x = 1 horizontally. A horizontal distance is simply the difference between the x-coordinates. The x-coordinates of all the points on the line x = 1 are 1's. Since we do not know if the x-coordinate of P is greater than or less than 1 we do not know which to put first in the difference. This problem can be solved using absolute value. This makes the distance from point P to the line x = 1:
{{{abs(x-1)}}}<br>
With the two expressions for the distances we can 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 the equation:
{{{sqrt(x^2-8x+16+y^2) = 2*abs(x-1)}}}<br>
Now we solve. Squaring both sides:
{{{x^2-8x+16+y^2 = (2*abs(x-1))^2}}}
{{{x^2-8x+16+y^2 = (2)^2*(abs(x-1))^2}}}
Squaring the 2 is simple. But how do you square an absolute value? The answer may be obvious to you. If not then a basic property of absolute values can help:
{{{abs(a)*abs(b) = abs(a*b)}}}
Now, if we rewrite {{{(abs(x-1))^2}}} without an exponent:
{{{abs(x-1)*abs(x-1)}}}
and then use the property we get:
{{{abs((x-1)(x-1))}}}
or
{{{abs((x-1)^2)}}}
Now we have the absolute value of a perfect square. But perfect squares can never be negative. And since the absolute value of a non-negative is simply itself, {{{abs((x-1)^2)}}} is just equal to:
{{{(x-1)^2}}}<br>
Back to squaring the right side of:
{{{x^2-8x+16+y^2 = (2)^2*(abs(x-1))^2}}}
which simplifies as follows:
{{{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)}}}<br>
Last of all we put it in general standard form. For this we gather all the terms on one side. Subtracting the entire right side from both sides we get:
{{{-3x^2+12+y^2 = 0}}}
Reordering:
{{{-3x^2+y^2+12 = 0}}}
Multiplying both sides by -1 (because I prefer positive leading coefficients):
{{{3x^2-y^2-12 = 0}}}<br>
At this point we should recognize the equation as the equation of a hyperbola. To help you graph it, let's put it in the standard form for hyperbolas:
{{{(x-h)^2/a^2 - (y-k)^2/b^2 = 1}}}
Adding 12 to both sides of our equation:
{{{3x^2-y^2=12}}}
Dividing both sides by 12:
{{{3x^2/12-y^2/12=1}}}
The first fraction reduces:
{{{x^2/4-y^2/12=1}}}
which can be rewritten as:
{{{(x-0)^2/4-(y-0)^2/12=1}}}<br>
From this we can see that the center is the point (0, 0), the {{{a^2 = 4}}} and the {{{b^2 = 12}}}. I'll leave the graph up to you.