Question

If the normal to the curve x = t – 1, y = 3t² – 6 at the point

(1, 6) make intercepts a and b on x and y-axis respectively, then the value of a + 12b is

• A. 132
• B. 146
• C. 150
• D. None of these

Answer 1

Answer: (B)

146

Answer 2

Given point is corresponding to t = 2 and

$\frac{dy}{dx}$ = 6t ̃ slope of normal at t = 2 is –$\frac{1}{12}$

\ equation of normal is y – 6 = – $\frac{1}{12}$ (x – 1)

̃ a = 73, b = $\frac{73}{12}$

̃ a + 12b = 146