2212
612
552
1572
Correct option is D. 552
Step 1: Use equation of normal in parametric form
Given equation of ellipse, 3x²+4y²=12 →x²/4+y²/3=1 →a²=4 →a=2 b²=3 →b=√3
Parametric coordinates of any point on an ellipse is (a cos θ, b sin θ)
Let point P(2cos θ, √3sin θ) (1)
Equation of parametric form of normal is ax sec θ - by csc θ = a² - b²
Equation of parametric form of normal at point P is 2x sec θ - √3y csc θ = 4 - 3 = 1
→2x sec θ - √3y csc θ = 1
Slope of normal = -2√3/3 * (cos θ/sin θ) = -2 tan θ/√3 (2)
Step 2: Find point P using given condition
Given that normal at P is parallel to the line, 2x+y=4
As we know parallel lines have same slope, So, slope of normal at P = Slope of line 2x+y=4
→Slope of normal at P = -2; From eq. (2)
-2 tan θ/√3 = -2; →tan θ = √3 →θ = π/3
Coordinates of point P = (2cos(π/3), √3sin(π/3)) (From eq (1))
→P = (21/2, √3√3/2) = (1, 3/2)
Step 3: Get the value of PQ using above result
P(1, 3/2) Q(4,4)
Distance between (x1,y1) (x2,y2) is √((x2-x1)²+(y2-y1)²)
→PQ = √((4-1)²+(4-3/2)²) = √(3²+(5/2)²) = √(9+25/4) = √(61/4) = √61/2 = 55/2
Hence, (D) is the correct option.