Consider a curve y=y(x) in the first quadrant as shown in the figure. Let the area A1 be twice the area A2. Then the normal to the curve perpendicular to the line 2x−12y=15 does NOT pass through the point:
Solutions · 1
Official solution
From the figure's setup, A1+A2=xy−8 and A1=2A2 give A1=32(xy−8), i.e. ∫4xf(t)dt=32(xf(x)−8). Differentiating:
\[f(x)=\frac23\left[xf'(x)+f(x)\right]\ \Longrightarrow\ \frac{f'(x)}{f(x)}=\frac1{2x}\ \Longrightarrow\ f^2(x)=cx.\]
Through (4,2): c=1, so the curve is y2=x.
Normal perpendicular to 2x−12y=15 (slope 1/6) has slope −6. On y2=x at (t2,t), curve slope =2t1, so normal slope =−2t=−6⇒t=3, point (9,3).
Normal: y=−6x+57. Checking the options, only (10,−4) fails (−6(10)+57=−3=−4).