Two problems on Conic Questions that are getting me stuck

[haree08]

a) The point (x,y) is equidistant from the circle x^2 +y^2 = 1 and the point (2,0). Show that (x,y) must lie on the curve (x-1)^2 +y^2 = 4(x-(5/4))^2. Show that this curve is a hyperbola.

b) If two tangents with slopes m1, m2 intersect at a point (X,Y) show that m1 and m2 must be the roots of the quadratic equation:
(a^2 - X) m^2 +2XY + (b^2 - Y^2) = 0
and deduce that if the tangents are perpendicular to each other, the point (X,Y) lies on a circle, centre the origin.

a) I am unsure how to show that the point MUST lie on the curve. All I can think of doing is finding a point that I know is equidistant between the curve and point, and then substituting these values into the equation of the curve? i.e. (1.5,0). Is this the correct way to approach this part of the problem.
To show that the curve is a hyperbola, do I just need to rearrange the equation they have given me into the form of a hyperbola ie. (x^2/a^2 - y^2/b^2 = 1) If so I have done this, and I have an equation that resembles a hyperbola, but I'm not sure this really SHOWS that the curve is a hyperbola. I think maybe I am missing something, any help would be great!

b) I don't know I am totally stuck!

 

[jvicens]

a) To show that the point (x,y) must lie on the curve, we can use the definition of equidistance. If a point is equidistant from two points, it must lie on the perpendicular bisector of the line segment connecting those two points. In this case, the two points are (2,0) and the point on the circle (x,y).

The midpoint of the line segment connecting these two points is (1,0). So, the equation of the perpendicular bisector is x = 1.

Now, we can substitute this value of x into the equation of the circle to get:

(1)^2 + y^2 = 1

Simplifying, we get y^2 = 0, which means y = 0.

So, the point (x,y) must have a y-coordinate of 0, which means it lies on the x-axis.

Substituting this into the equation of the circle, we get:

x^2 + (0)^2 = 1

Simplifying, we get x = ±1.

So, the point (x,y) must have an x-coordinate of ±1, which means it lies on the line x = ±1.

Substituting these values into the equation of the curve given, we get:

(x-1)^2 + (0)^2 = 4(x-(5/4))^2

Simplifying, we get x^2 - 2x + 1 = 4x^2 - 20x + 25

Rearranging, we get 3x^2 - 22x + 24 = 0

This is the equation of a hyperbola, since the coefficient of x^2 is positive.

b) To show that m1 and m2 must be the roots of the quadratic equation, we can use the fact that they are the slopes of the tangents at the point (X,Y).

The equation of a tangent to a curve at a point (X,Y) can be written as:

y - Y = m(x - X)

Substituting the slopes m1 and m2, we get:

y - Y = m1(x - X) and

y - Y = m2(x - X)

Solving these two equations simultaneously, we get:

m1x - m2x