Question

The slope of the normal to the curve x = t² + 3t – 8, y = 2t² – 2t – 5 at the point (–10, –1) is

वक्र x = t² + 3t – 8, y = 2t² – 2t – 5 के बिंदु (–10, –1) पर अभिलम्ब की प्रवणता है

• A. –6
• B. $-\frac{1}{6}$
• C. $\frac{1}{6}$
• D. 6

Answer 1

The correct option is C $\frac{1}{6}$

Solutions