The slope of the curve 2y2 = ax2 + b at (1, – 1) is – 1. Find a, b

Asked by Topperlearning User | 7th Aug, 2014, 08:32: AM

Expert Answer:

The equation of the curve is

2y2 = ax2 + b    …(i)
Differentiating w.r.t. x, we get
4 straight y. dy over dx equals 2 ax 
rightwards double arrow dy over dx equals fraction numerator ax over denominator 2 straight y end fraction rightwards double arrow open parentheses dy over dx close parentheses subscript left parenthesis 1 comma minus 1 right parenthesis end subscript equals minus straight a over 2
It is given that the slope of the tangent at (1, –1) is –1. Therefore,
minus straight a over 2 equals minus 1 rightwards double arrow straight a equals 2
Since the point (1, –1) lies on (i). Therefore,
2 (–1)2 = a (1)2 + b Þ a + b = 2
Putting a = 2 in a + b = 2, we obtain b = 0.
Hence a = 2 and b = 0.

Answered by  | 7th Aug, 2014, 10:32: AM