Consider a curve y = y x in the first quadrant as shown in the figure. Let the area A 1 is twice the area A 2 . Then the normal to the curve perpendicular to the line 2 x - 12 y = 15 does NOT pass through the point __ ​​​​​​​

Step-by-step solution

Given that A 1 = 2 A 2 Now, from the graph A 1 + A 2 = x × y - 2 × 4 = x y - 8 ⇒ 3 2   A 1 = x y - 8 ⇒ A 1 = 2 3 x y - 16 3 ⇒ ∫ 4 x f x d x = 2 3 x y - 16 3 ⇒ f x = 2 3 x d y d x + y ⇒ 2 3 x d y d x = y 3 ⇒ 2 ∫ d y y = ∫ d x x ⇒ 2 ln y = ln x + ln c ⇒ y 2 = c x As f 4 = 2 ⇒ c = 1 So y 2 = x is the given curve. Given that the slope of normal = - 6 So the equation of normal will be y = - 6 x - 1 2 - 6 - 1 4 - 6 3 ⇒ y = - 6 x