Application of Derivatives

Question

For the curve y = 4x³ – 2x⁵, find all the points at which the tangent passes through the origin.

Solution

The equation of curve is
y = 4x³ – 2x^{5 ...(1)}Let the tangent to the curve (1) at (h, k) pass through origin (0, 0)
   $therefore space space space straight k space equals space 4 straight h cubed minus 2 straight h to the power of 5$ ...(2)
$open square brackets because space space space space left parenthesis straight h comma space straight k right parenthesis space lies space on space left parenthesis 1 right parenthesis close square brackets$
Differentiating (1) w.r.t.x, $dy over dx space equals space 12 straight x squared minus 10 straight x to the power of 4$
$At space left parenthesis straight h comma space straight k right parenthesis comma space space dy over dx space equals space 12 straight h squared minus 10 straight h to the power of 4 comma space space which space is space slope space of space tangent.$
The equation of tangent of (h, k) is
   $straight y minus straight k space equals space left parenthesis 12 straight h squared minus 10 straight h to the power of 4 right parenthesis thin space left parenthesis straight x minus straight h right parenthesis$
or $straight y equals left parenthesis 4 straight h cubed minus 2 straight h to the power of 5 right parenthesis space equals space left parenthesis 12 straight h squared minus 10 straight h to the power of 4 right parenthesis thin space left parenthesis straight x minus straight h right parenthesis space space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets$
Now this tangent passes through (0, 0).
$therefore space space space space space space 0 minus left parenthesis 4 straight h cubed minus 2 straight h to the power of 5 right parenthesis space equals space left parenthesis 12 straight h squared minus 10 straight h to the power of 4 right parenthesis thin space left parenthesis 0 minus straight h right parenthesis space space space space space space space space$
or    $negative 4 straight h cubed plus 2 straight h to the power of 5 space equals space minus 12 straight h cubed plus 10 straight h to the power of 5 space space space space or space space space space 8 straight h to the power of 5 minus 8 straight h cubed space equals space 0$
or    $straight h to the power of 5 minus straight h cubed space equals space 0 space space space space space or space space space straight h cubed left parenthesis straight h squared minus 1 right parenthesis space equals space 0$ .
$rightwards double arrow$    $straight h space equals space 0 comma space space space 1 comma space space space minus 1$
$therefore space space space space space straight k space equals space 0 comma space space 4 minus 2 comma space space minus 4 plus 2 space space space space space space space space space space space space space space space space space space space space space space space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets space space space space space space space space space space space space space equals space 0 comma space space 2 comma space space minus 2 therefore space required space points space are space left parenthesis 0 comma space 0 right parenthesis comma space left parenthesis 1 comma space 2 right parenthesis comma space left parenthesis negative 1 comma space minus 2 right parenthesis$

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