`star` Tangents And Normals

`\color{green} ✍️` 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

`color{red}{(dy)/(dx) ]_(x_0, y_0) = f' (x_0}`. So

`\color{green} ✍️` So The equation of the tangent at `(x_0, y_0)` to the curve y = f (x) is given by

`color{green}{y – y_0 = f ′(x_0)(x – x_0)}`

`=>`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 Normal to the curve` y = f (x)` at `(x_0, y_0) ` is given by

`color{green}{y - y_0 = (-1)/( f' (x_0 ) ) (x - x_0)}`

i.e., `color{green}{ (y- y_0) f' (x_0 ) + (x - x_0 ) = 0}`

`color{red}{"Key Concept :"}` If a tangent line to the curve `y = f (x)` makes an angle `θ` with `x-`axis in the positive direction, then

`color{blue}{(dy)/(dx) = "slope of the tangent" = tan θ}`

`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

`color{red}{(dy)/(dx) ]_(x_0, y_0) = f' (x_0}`. So

`\color{green} ✍️` So The equation of the tangent at `(x_0, y_0)` to the curve y = f (x) is given by

`color{green}{y – y_0 = f ′(x_0)(x – x_0)}`

`=>`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 Normal to the curve` y = f (x)` at `(x_0, y_0) ` is given by

`color{green}{y - y_0 = (-1)/( f' (x_0 ) ) (x - x_0)}`

i.e., `color{green}{ (y- y_0) f' (x_0 ) + (x - x_0 ) = 0}`

`color{red}{"Key Concept :"}` If a tangent line to the curve `y = f (x)` makes an angle `θ` with `x-`axis in the positive direction, then

`color{blue}{(dy)/(dx) = "slope of the tangent" = tan θ}`

(i)`\color{green} ✍️` 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 `color{green}{y = y_0}`.

(ii) If `θ -> 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 `color{green}{x = x_0}`.

In this case, the equation of the tangent at the point `(x_0, y_0)` is given by `color{green}{y = y_0}`.

(ii) If `θ -> 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 `color{green}{x = x_0}`.

Q 3115345269

Find the slope of the tangent to the curve `y = x^3 – x` at x = 2.

Class 12 Chapter 6 Example 14

Class 12 Chapter 6 Example 14

The slope of the tangent at x = 2 is given by

` (dy)/(dx) ]_(x=2) = 3x^2 -1 ]_(x=2) = 11` .

Q 3125445361

Find the point at which the tangent to the curve `y = sqrt (4x-3) -1` has its slope `2/3`

Class 12 Chapter 6 Example 15

Class 12 Chapter 6 Example 15

Slope of tangent to the given curve at (x, y) is

` (dy)/(dx) = 1/2 (4x-3)^(-1/2) 4 = 2/(sqrt (4x-3) )`

The slope is given to be `2/3`

So `2/(sqrt (4x-3) ) = 2/3`

or `4x-3 = 9`

or `x= 3`

Now `y = sqrt (4x-3) -1` . So when `x = 3, y = sqrt (4(3) -3) -1= 2` .

Therefore, the required point is (3, 2).

Q 3155445364

Find the equation of all lines having slope 2 and being tangent to the curve

`y + 2/(x-3) = 0`

Class 12 Chapter 6 Example 16

`y + 2/(x-3) = 0`

Class 12 Chapter 6 Example 16

Slope of the tangent to the given curve at any point (x,y) is given by

` (dy)/(dx) = 2/((x-3)^2)`

But the slope is given to be 2. Therefore

`2/((x-3)^2) = 2`

or ` (x-3)^2 = 1`

or ` x-3 = pm 1`

or `x= 2,4`

Now x = 2 gives y = 2 and x = 4 gives y = – 2. Thus, there are two tangents to the

given curve with slope 2 and passing through the points (2, 2) and (4, – 2). The equation

of tangent through (2, 2) is given by

y – 2 = 2(x – 2)

or y – 2x + 2 = 0

and the equation of the tangent through (4, – 2) is given by

y – (– 2) = 2(x – 4)

or y – 2x + 10 = 0

Q 3155845764

Find the equation of the tangent to the curve ` y = (x-7)/( ( x-2)(x-3) )` at the point where it cuts the x-axis.

Class 12 Chapter 6 Example 18

Class 12 Chapter 6 Example 18

Note that on x-axis, y = 0. So the equation of the curve, when y = 0, gives

x = 7. Thus, the curve cuts the x-axis at (7, 0). Now differentiating the equation of the

curve with respect to x, we obtain

` (dy)/(dx) = (1-y (2x-5) )/( (x-2)(x-3) ) ` (Why?)

or ` (dy)/(dx)]_(7,0) = (1-0)/((5)(4)) = 1/20`

Therefore, the slope of the tangent at (7, 0) is `1/20` . Hence, the equation of the

tangent at (7, 0) is

` y- 0 = 1/20 (x-7)` or `20y − x + 7 = 0`

Q 3125845761

Find points on the curve ` x^2/4 + y^2/25 =1` at which the tangents are (i) parallel to x-axis (ii) parallel to y-axis.

Class 12 Chapter 6 Example 17

Class 12 Chapter 6 Example 17

Differentiating `x^2/4 + y^2/25 =1` with respect to x, we get

`x/2 + (2y)/(25) (dy)/(dx) = 0`

or ` (dy)/(dx) = (-25)/4 x/y`

(i) Now, the tangent is parallel to the x-axis if the slope of the tangent is zero which

gives ` (-25)/4 x/y = 0` . This is possible if x = 0. Then `x^2/4 + y^2/25 =1` for x = 0 gives

`y^2 =25` i.e., `y = pm 5`

Thus, the points at which the tangents are parallel to the x-axis are (0, 5) and

(0, – 5).

(ii) The tangent line is parallel to y-axis if the slope of the normal is 0 which gives

` (4y)/(25 x) = 0` , i.e. `y =0` Therefore , `x^2/4 + y^2/25 = 1` for y = 0 gives x = ± 2. Hence, the

points at which the tangents are parallel to the y-axis are (2, 0) and (–2, 0).

Q 3175845766

Find the equations of the tangent and normal to the curve `x^2/3 + y^2/3 = 2` at `(1,1)`

Class 12 Chapter 6 Example 19

Class 12 Chapter 6 Example 19

Differentiating `x^(2/3) + y^(2/3) = 2` with respect to x, we get

`2/3 x^(-1/3) + 2/3 y^(-1/3) (dy)/(dx) = 0`

or ` (dy)/(dx) = - (y/x)^1/3`

Therefore, the slope of the tangent at (1, 1) is ` (dy)/(dx)]_(1,1) = -1`

So the equation of the tangent at (1, 1) is

y – 1 = – 1 (x – 1) or y + x – 2 = 0

Also, the slope of the normal at (1, 1) is given by

`(-1)/( text (slope of the tangent at (1,1)) ) = 1`

Therefore, the equation of the normal at (1, 1) is

y – 1 = 1 (x – 1) or y – x = 0

Q 3105845768

Find the equation of tangent to the curve given by

`x = a sin^3 t , y = b cos^3 t` ... (1)

at a point where ` t = pi/2` .

Class 12 Chapter 6 Example 20

`x = a sin^3 t , y = b cos^3 t` ... (1)

at a point where ` t = pi/2` .

Class 12 Chapter 6 Example 20

Differentiating (1) with respect to t, we get

`(dx)/(dt) = 3 a sin^2 t cos t ` and ` (dy)/(dt) = -3 b cos^2 t sin t`

or ` (dy)/(dx) = ( (dy)/(dt) )/( (dx)/(dt) ) = (-3 b cos^2 t sin t)/(3 a sin^2 t cos t ) = (-b)/a (cos t )/( sin t )`

Therefore, slope of the tangent at ` t = pi/2` i

`(dy)/(dx)]_(t = pi/2) = (-b cos pi/2 )/( a sin pi/2) =0`

Also, when ` t = pi/2 , x = a` and `y = 0` . Hence, the equation of tangent to the given

curve at `t = pi/2` , i.e. at ` (a, 0)` is

y – 0 = 0 (x – a), i.e., y = 0.