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The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is: (A) A/2a (B) A/a (C) 2A/a (D) 4A/a Correct answer is (C). PLZ EXPLAIN IT.?
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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
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The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such...
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The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is: (A) A/2a (B) A/a (C) 2A/a (D) 4A/a Correct answer is (C). PLZ EXPLAIN IT.?
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The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is: (A) A/2a (B) A/a (C) 2A/a (D) 4A/a Correct answer is (C). PLZ EXPLAIN IT.? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is: (A) A/2a (B) A/a (C) 2A/a (D) 4A/a Correct answer is (C). PLZ EXPLAIN IT.? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The triangle PQR of area 'A' is inscribed in the parabola y^2=4ax such that the vertex P lies at the vertex of the parabola and the base QR is a focal chord. The modulus of the difference of the ordinates of the points Q and R is: (A) A/2a (B) A/a (C) 2A/a (D) 4A/a Correct answer is (C). PLZ EXPLAIN IT.?.
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