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

Fig. 

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

Answer 1

The Correct Option is C

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)