Let P and Q be two points on a parabola whose vertex is O. If the chord PQ meets the axis at K and the tangent to the parabola at P and Q meet the axis at T1 and T2. Prove that OT1.OT2=OK^2
To prove that OT1 * OT2 = OK^2, we need to use the properties of the parabola and the given points.
Let's start with a brief explanation of the properties involved:
- Vertex (O): The vertex of a parabola is the point where the parabola reaches its maximum or minimum point.
- Chord (PQ): A chord is a line segment that connects two points on a curve (in this case, the two points are P and Q on the parabola).
- Axis: The axis of a parabola is a vertical line passing through the vertex, denoted as the x-axis in our case.
- Tangent: A tangent is a line that touches the curve at a single point (in this case, tangents to the parabola are drawn at points P and Q).
Now, let's prove the given statement.
Since P and Q are two points on the parabola, the tangent lines at these points intersect the axis at T1 and T2, respectively. Let's denote the coordinates of P as (x1, y1) and Q as (x2, y2).
To find the equation of the parabola, we need three points. We have the vertex O, point P, and point Q.
Since the vertex is given as O, we know its coordinates are (0, 0) because the vertex lies on the axis.
Now we have three points, O(0, 0), P(x1, y1), and Q(x2, y2). We can use these points to find the equation of the parabola.
The equation of a parabola in its standard form is y^2 = 4ax, where a is a constant specific to each parabola.
Using the vertex form of a parabola, we can rewrite the equation as:
(y - k)^2 = 4a(x - h), where (h, k) represents the vertex.
In our case, we substitute (h, k) = (0, 0) and solve for a using the coordinates of point P or Q.
We can assume that P is the point (-p, 2ap) where p is the distance between the vertex and point P in the x-direction. Substituting these coordinates into the equation, we have:
(2ap)^2 = 4a(-p)
Simplifying, we get:
4a^2p^2 = -4ap
Dividing both sides by 4a and cancelling p, we have:
ap = -1
Now we have the value of a.
Next, we need to find the coordinates of points T1 and T2 where the tangents intersect the axis. The x-coordinate of T1, denoted by xT1, is equal to -2a multiplied by the y-coordinate of P, which is -2ap. Therefore, xT1 = -2a * (-2ap) = 4a^2p.
Similarly, the x-coordinate of T2, denoted by xT2, is equal to 4a^2p as well.
Now, we have the coordinates of T1 and T2 as (4a^2p, 0) and (-4a^2p, 0) respectively.
Notice that the lengths OT1 and OT2 are equal to 2ap, as the distance from the origin O to the x-coordinate of T1 (or T2) is 2ap.
Now, we need to find the length of the chord PQ, which is given by the distance formula:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Substituting the coordinates of P and Q into the formula, we get:
d = sqrt((x2 - x1)^2 + (2ap - 2ap)^2)
= sqrt((x2 - x1)^2)
= |x2 - x1|
Since P and Q lie on the parabola, the distance between them is equal to the x-coordinate difference: |x2 - x1|
Since the chord PQ intersects the axis at point K, the x-coordinate of K, denoted by xK, is equal to the average of x1 and x2. Therefore, xK = (x1 + x2)/2.
Now, we can express the length of the chord as:
d = |x2 - x1| = |2xK - (x1 + x2)| = |2xK - 2(xK)| = 2|xK|
Since OT1 = OT2 = 2ap, and OK = |xK|, we can rewrite our initial equation as:
OT1 * OT2 = (2ap)(2ap) = 4a^2p^2
Using our previous result, ap = -1, we get:
4a^2p^2 = 4a^2(-1)^2 = 4a^2 = (2a)^2 = OK^2
Therefore, we have proved that OT1 * OT2 = OK^2.