Find the points on the curve 9y2 = x3 where normal to the curve makes equal intercepts with the axes.

Asked by Topperlearning User | 7th Aug, 2014, 12:43: PM

Expert Answer:

Let the required point be (x1, y1)

The equation of the curve is 9y2 = x2
Since (x1, y1) lies on the curve. Therefore,
9 straight y subscript 1 squared equals straight x subscript 1 cubed...... left parenthesis straight i right parenthesis 
Now, 9y2 = x3
 

rightwards double arrow dy over dx equals fraction numerator straight x squared over denominator 6 straight y end fraction
rightwards double arrow open parentheses dy over dx close parentheses subscript left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis end subscript equals fraction numerator straight x subscript 1 squared over denominator 6 straight y subscript 1 end fraction

 
Since the normal to the curve at (x1, y1) make equal intercepts with the coordinate axes. Therefore,
Slope of the normal = ± 1
 

rightwards double arrow minus 1 over open parentheses begin display style dy over dx end style close parentheses subscript left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis end subscript equals plus-or-minus 1
rightwards double arrow open parentheses dy over dx close parentheses subscript left parenthesis straight x subscript 1 comma straight y subscript 1 right parenthesis end subscript equals plus-or-minus 1
rightwards double arrow fraction numerator straight x subscript 1 squared over denominator 6 straight y subscript 1 end fraction equals plus-or-minus 1
rightwards double arrow straight x subscript 1 to the power of 4 equals 36 straight y subscript 1 squared
rightwards double arrow straight x subscript 1 to the power of 4 equals 36 open parentheses straight x subscript 1 cubed over 9 close parentheses left square bracket using space left parenthesis straight i right parenthesis right square bracket
rightwards double arrow straight x subscript 1 to the power of 4 equals 4 straight x subscript 1 cubed
rightwards double arrow straight x subscript 1 cubed left parenthesis straight x subscript 1 minus 4 right parenthesis equals 0
rightwards double arrow straight x subscript 1 equals 0 comma space 4

 
Putting x1 = 0 in (i), we get
9 straight y subscript 1 squared equals 0 rightwards double arrow straight y subscript 1 equals 0
Putting x1 = 4 in (i), we get
9 straight y subscript 1 squared equals 4 cubed rightwards double arrow straight y subscript 1 equals plus-or-minus 8 over 3
But, the line making equal intercepts with the coordinate axes cannot pass through the origin.
Hence, the required points are open parentheses 4 comma 8 over 3 close parentheses space & space open parentheses 4 comma minus 8 over 3 close parentheses

Answered by  | 7th Aug, 2014, 02:43: PM