The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such...
Solution:
Step 1: Finding the equation of QR
Let the coordinates of Q be (at², 2at) and the coordinates of R be (a/u², -2au). Since QR is a focal chord, it passes through the focus (a, 0). Hence, the equation of QR is
y = m(x-a)
where m is the slope of QR. Substituting the coordinates of Q and R, we get
2at = m(at² - a) ...(1)
-2au = m(a/u² - a) ...(2)
Solving equations (1) and (2) for m, we get
m = -2a/(t+u)
Substituting this value of m in the equation of QR, we get
y = -2a(x-a)/(t+u)
Simplifying, we get
xt + uy - 2a(t+u) = 0 ...(3)
Step 2: Finding the area of triangle PQR
The vertex P lies at the vertex of the parabola, which is (0, 0). Hence, the coordinates of P are (0, 0). The area of triangle PQR is given by
A = (1/2) |x1(y2-y3) + x2(y3-y1) + x3(y1-y2)|
where (x1, y1) = (0, 0), (x2, y2) = (at², 2at) and (x3, y3) = (a/u², -2au). Substituting these values, we get
A = (1/2) |0 - 4a³t/u - 4a³u/t|
Simplifying, we get
A = 2a³ |t-u|/tu
Step 3: Finding |t-u|
Substituting equation (3) in the equation of the parabola, we get
y² = 4ax
Substituting y = -2au/(t+u) and x = a(tu)/(t+u), we get
(-2au/(t+u))² = 4a(a(tu)/(t+u))
Simplifying, we get
(t-u)² = 4tu
Taking the modulus, we get
|t-u| = 2√(tu)
Substituting this value in the expression for A, we get
A = 4a³√(tu)
Dividing by 2a, we get
A/2a = 2a²√(tu)/2a
Simplifying, we get
A/2a = a√(tu)
Squaring both sides, we get
(A/2a)² = a²tu
Dividing by A, we get
(A/2a)²/A = a²tu/A
Simplifying, we get
(A/2a)/A = |t-u|/4a
Substituting the value of |t-u|, we get
(A/2a)/A = √(tu)/2a
Multiplying by 2, we get
2(A/2