Consider a curve y = y(x) in the first quadrant as shown in the figure.

Let the area A₁ is twice the area A₂. Then the normal to the curve perpendicular to the line 2x − 12y = 15 does NOT pass through the point.

Calculation

Given that A₁ = 2A₂

from the graph A₁ + A₂ = xy - 8

⇒ \(\frac{3}{2}\) A₁ = xy - 8

⇒ A₁ = \(\frac{2}{3}\) xy - \(\frac{16}{3}\)

⇒ \(\int_4^x f(x) dx = \frac{2}{3}xy - \frac{16}{3}\)

⇒ f(x) = \(\frac{2}{3}(x\frac{dy}{dx} + y)\)

⇒ \(\frac{2}{3}x\frac{dy}{dx} = \frac{y}{3}\)

⇒ \(2\int\frac{dy}{y} = \int\frac{dx}{x}\)

⇒ 2 ln y = ln x + ln c

⇒ y² = cx

As f(4) = 2 ⇒ c = 1

so y² = x

slope of normal = -6

y = -6(x) - \(\frac{1}{2}\)(-6) - \(\frac{1}{4}\)(-6)³

y = -6x + 3 + 54

⇒ y + 6x = 57

(3) not satisfy the equation

Hence option 3 is correct