CBSE Class 12-science Answered
Find the point on the curve y = x3 – 11x + 5 at which the tangent has the equation y = x – 11.
Asked by Topperlearning User | 11 Aug, 2014, 10:06: AM
Let the required point be P(x1, y1), since (x1, y1) lies on y = x3 – 11x + 5.
y1 =
…(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](https://images.topperlearning.com/topper/tinymce/cache/e56a630977f1cf277d8a03656c3610a1.png)
![x subscript 1 cubed minus 11 x subscript 1 plus 5](https://images.topperlearning.com/topper/tinymce/cache/b8f6309827d1a726bb1ba607a1845d75.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/e56a630977f1cf277d8a03656c3610a1.png)
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](https://images.topperlearning.com/topper/tinymce/cache/592c65095a3bff4863ce9a6dd6a600fd.png)
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](https://images.topperlearning.com/topper/tinymce/cache/650ea77f714abe8b181ede99fbbf51cb.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/7d2a78f948dbba81a80d3de3c85b4d49.png)
Now,
[Using (i)]
![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](https://images.topperlearning.com/topper/tinymce/cache/bb0224a36e2b18bb209b3470585be206.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/4e094b90465efce5464d65b25844743d.png)
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 | 11 Aug, 2014, 12:06: PM
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