Q. 14 A3.6( 39 Votes )
Find the equations of the tangent and normal to the given curves at the indicated points:
y = x4 – 6x3 + 13x2 – 10x + 5 at (0, 5)
Answer :
It is given that equation of curve is y = x4 – 6x3 + 13x2 – 10x + 5
On differentiating with respect to x, we get
= 4x3 - 18x2 +26x -10
Now, Slope of tangent will be the value of
at x = 0 i.e. m1 = 4(0)3 - 18(0)2 + 26(0) - 10 = -10
Therefore, the slope of the tangent at (0, 5) is -10.
Then, the equation of the tangent is
y – 5 = -10(x – 0)
⇒ y – 5 = 10x
⇒ 10x +y =5
Also, slope of normal at (0,5)
Now, equation of the normal at (0,5)
⇒ 10y - 50 = x
⇒ x -10y +50 = 0
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