Question

The point of the curve $y^2=2(x-3)$ at which the normal is parallel to the line y - 2x + 1 = 0 is
[MP PET 1998]

• A. (5, 2)
• B. $(-\frac{1}{2},-2)$
• C. (5, -2)
• D. $(\frac{3}{2},2)$

Answer 1

The correct option is D $(\frac{3}{2},2)$

Let $y=x^{6}andz=x^3;\therefore \frac{dy}{dz}=\frac{dy/dx}{dz/dx}=\frac{6x^5}{3x^2}=2x^3.$