Maharashtra State BoardHSC Arts 12th Board Exam
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Tangents and Normals

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notes

The equation of a straight line passing through a given point `(x_0, y_0)` having finite slope m is given by 
y - `y_0` = m (x - `x_0`)
Note that the slope of the tangent to the curve y = f(x) at the point `(x_0, y_0)` is given by `(dy)/(dx)]_(x_0,y_0)` `(=f'(x_0))`.
So the equation of the tangent at `(x_0,y_0)` to the curve y = f(x) is given by `y-y_0 = f'(x_0)(x - x_0)`Fig.

Also, since the normal is perpendicular to the tangent, the slope of the normal to the curve y = f(x) at  `(x_0, y_0)` is `(-1)/(f'(x_0))` , if  f'(x_0) ≠ 0 . Therefore, the equation of the normal to the curve y = f(x)  at `(x_0, y_0)`  is given by

`y – y_0 = (-1)/(f'(x_0))(x - x_0)`

i.e. `(y-y_0) f'(x_0) + (x-x_0) = 0`

Particular cases: 

(i) If slope of the tangent line is zero, then tan θ = 0 and so θ = 0 which means the tangent line is parallel to the x-axis. In this case, the equation of the tangent at the point `(x_0, y_0)` is given by `y = y_0`.

(ii) If `theta →pi / 2 `, then tan θ → ∞, which means the tangent line is perpendicular to the x-axis, i.e., parallel to the y-axis. In this case, the equation of the tangent at `(x_0, y_0)`  is given by `x = x_0`

Video link : https://youtu.be/Rf-h-wktw-s

Video Tutorials

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Shaalaa.com | Tangents and Normals part 1

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Tangents and Normals part 1 [00:27:13]
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