Question
Question: Find the length of the chord of contact of the tangents drawn from the point \(\left( {{x}_{1}},{{y}...
Find the length of the chord of contact of the tangents drawn from the point (x1,y1) to the parabola y2=4ax
Solution
We need to find the length of the tangents to the parabola y2=4ax . Let us consider that the tangents at points P and Q meet at (x1,y1) . We start to solve the question by finding out the value of the coordinates (x1,y1) . Then, we find the distance between the points P, Q to get the desired result.
Complete step-by-step answer:
We are given an equation of a parabola and need to find the length of chord of contact of the tangents. We start to solve the question by finding out the value of coordinates (x1,y1) and then find the length of the chord of contact of the tangents.
A line touching a parabola is said to be a tangent to a parabola. The tangent to a parabola touches it at exactly one point.
The chord that joins the point of contact of the tangents from an external point is called the chord of contact of a parabola.
The given question can be diagrammatically represented as follows,
From the above figure,
The tangents to the parabola meet at the points A,B on the parabola.
The points A,B are given as follows,
A=(at12,2at1) ;
B=(at22,2at2)
The tangents at the points A,B meets at a point (x1,y1)
From the above,
The coordinates of (x1,y1) are given as follows,
x1=at1t2
y1=a(t1+t2)
The length of the chord of contact of the tangents drawn from the point (x1,y1) to the parabola y2=4ax is the distance between the points A,B .
We know that distance between two points (x1,y1),(x2,y2) is given by the formula
=(x1−x2)2+(y1−y2)2
Following the same, the distance between the points A(at12,2at1),B(at22,2at2) is given by
=(at12−at22)2+(2at1−2at2)2
Simplifying the above equation, we get,
=a2(t12−t22)2+4a2(t1−t2)2
Taking a2 from the square root, we get,
=a(t12−t22)2+2(t1−t2)2
Simplifying the above equation by expanding the terms, we get,
=a((t1+t2)2−4t1t2)((t1+t2)2+4)
Substituting the values with x1,y1 in the above equation, we get,
=a((ay1)2−4(ax1))((ay1)2+4)
Simplifying the above equation, we get,
=a(a2y12−4ax1)(a2y12+4a2)
Taking a2 out of the square root, we get,
=aaa2(y12−4ax1)(y12+4a2)
=a2(y12−4ax1)(y12+4a2)
Note: We need to know that distance between two points (x1,y1),(x2,y2) is given by the formula
=(x1−x2)2+(y1−y2)2
The values of the coordinates (x1,y1) on the parabola y2=4ax is given by x1=at1t2
y1=a(t1+t2) such that they are the point of intersection of both the tangents to the parabola.