Question -
Answer -
(i) The equation of the curve is y = x4 −6x3 + 13x2 − 10x +5.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (0, 5) is −10. Theequation of the tangent is given as:
y − 5 =− 10(x − 0)
⇒ y − 5 = − 10x
⇒ 10x + y = 5
The slope of the normal at (0, 5) is 
Therefore, the equation of the normal at (0, 5) is givenas:
(ii) The equation of the curve is y = x4 −6x3 + 13x2 − 10x +5.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (1, 3) is 2. Theequation of the tangent is given as:
The slope of the normal at(1, 3) is
Therefore, the equation of the normal at (1, 3) is givenas:
(iii) The equation of the curve is y = x3.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (1, 1) is 3 and theequation of the tangent is given as:
The slope of the normal at(1, 1) is
Therefore, the equation of the normal at (1, 1) is givenas:
(iv) The equation of the curve is y = x2.
On differentiating with respect to x,we get:
Thus, the slope of the tangent at (0, 0) is 0 and theequation of the tangent is given as:
y − 0 =0 (x − 0)
⇒ y = 0
The slope of the normal at (0, 0) is
, which is not defined.
Therefore, the equation of the normal at (x0, y0) = (0, 0) is given by
(v) The equation of the curve is x =cos t, y = sin t.
∴The slope of the tangent at
is −1.
When
Thus, the equation of thetangent to the given curve at 
The slope of the normal atis 
Therefore, the equation ofthe normal to the given curve at