Find the point on the curve y = x3 – 11x + 5 at which the tangent has the equation y = x – 11.

Asked by Topperlearning User | 11th Aug, 2014, 10:06: AM

Expert Answer:

Let the required point be P(x1, y1), since (x1, y1) lies on y = x3 – 11x + 5.

y1 = x subscript 1 cubed minus 11 x subscript 1 plus 5  …(i)open parentheses fraction numerator d y over denominator d x end fraction close parentheses subscript left parenthesis x subscript 1 comma y subscript 1 right parenthesis end subscript equals
Now, y = x3 – 11x + 5
 straight y equals straight x cubed minus 11 straight x plus 5
rightwards double arrow dy over dx equals 3 straight x squared minus 11
rightwards double arrow open parentheses dy over dx close parentheses subscript open parentheses straight x subscript 1 comma straight y subscript 1 close parentheses end subscript equals 3 straight x subscript 1 squared minus 11 
 
Since the line y = x – 11 is tangent at the point (x1, y1). Therefore,
Slope of the tangent at (x1, y1) = (Slope of the line y = x – 11).
open parentheses dy over dx close parentheses subscript open parentheses straight x subscript 1 comma straight y subscript 1 close parentheses end subscript equals(Slope of the line x – y – 11 = 0)
rightwards double arrow 3 straight x subscript 1 squared minus 11 equals fraction numerator minus 1 over denominator minus 1 end fraction open square brackets because Slope space equals minus fraction numerator coeff. space of space straight x over denominator coeff. space of space straight y end fraction close square brackets
rightwards double arrow 3 straight x subscript 1 squared equals 12 rightwards double arrow straight x subscript 1 squared equals 4 rightwards double arrow straight x subscript 1 equals plus-or-minus 2
Now,x subscript 1 equals 2 rightwards double arrow y equals 2 cubed minus 11 left parenthesis 2 right parenthesis plus 5 rightwards double arrow y equals minus 9 [Using (i)]
straight x subscript 1 equals minus 2 rightwards double arrow straight y equals left parenthesis minus 2 right parenthesis cubed minus 11 left parenthesis minus 2 right parenthesis plus 5 rightwards double arrow straight y equals 19 [Using (i)]
So, two points are (2, –9) and (–2, 19). Of these two points (–2, 19) does not lie on y = x – 11. Therefore, the required point is (2, – 9).

Answered by  | 11th Aug, 2014, 12:06: PM