Question

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 – 12 y = 15 does NOT pass through the point.

Fig.

• A. (6,21)
• B. (8,9)
• C. (10,-4)
• D. (12,-15)

Answer 1

Answer: (C)

(10,-4)

Answer 2

The correct answer is (C) : (10,-4)
$A_1+A_2 = xy-8\ \&\ A_1 = 2A_2$
$A_1+\frac{A_1}{2} = xy-8$
$A_1 = \frac{2}{3}(xy-8)$
$\int_{4}^{x} f(x) \, dx = \frac{2}{3} (x f(x) - 8)$
Differentiate w.r.t.x
f(x) = \frac{2}{3} \left\\{xf'(x)+f(x)\right\\}
$\frac{2}{3}xf'(x) =\frac{1}{3} f(x)$
$2∫\frac{f'(x)}{f(x)}dx = ∫\frac{dx}{x}$
$2lnf(x) = lnx+lnc$
$f^2(x) = cx$
Which passes through (4, 2)
$4 = c×4 ⇒ c = 1$
Equation of required curve y2 = x
Equation of normal having slope (–6) is
$y = -6x - 2(\frac{1}{4})(-6)-\frac{1}{4}(-6)^3$
$y = -6x + 57$
Which does not pass through (10, –4)