Question

Find the equation of the tangent to the curve x^2+3y-3=0

Answer 1

Answer:

3 y + 2 x = 3 - x₁² + 2 x₁

Step-by-step explanation:

Given curve x^2 + 3 y - 3 = 0  --- (1)

Differentiate with respect to x for getting the slope of the tangent at point (x,y).

So 2 x + 3 y' = 0

So y' = - 2/3 x    --- (2)

Thus the tangent to the curve at a point P (x₁, y₁) will be :

    y = -2/3 x₁ + c   or   3 y + 2 x = c   --- (3)

As P is on curve (1),  x₁² + 3 y₁ = 3   or  3 y₁ = 3 - x₁²

As P is on the tangent too,

  3 - x₁² + 2 x₁ = c  

Finally the equation of the tangent  at P(x₁, y₁) :

               3 y + 2 x = 3 - x₁² + 2 x₁

Answer 2

Step 1:

Given curve is x2+3y=3x2+3y=3

Let y=−x2+33y=−x2+33

On differentiating with respect to xx

∴dydx=13∴dydx=13[−2x][−2x]

=−23=−23xx

Step 2:

Since the tangent is parallel to the line y−4x+5=0y−4x+5=0

Then slopes should be equal

Slope of the given line is 4

−2x3=−2x3=44

x=−4×32x=−4×32

=−122=−122

=−6=−6

Step 3:

∴y=−(−6)2+33∴y=−(−6)2+33

=−36+33=−36+33

=−333=−333

=−11=−11

Hence the points of contact are (−6,−11)(−6,−11)

Step 4:

Equation of the tangent at (x1,y1)(x1,y1) where slope is mm is given by y−y1=m(x−x1)y−y1=m(x−x1)

(i.e)[y−(−11)]=4(x−(−6))[y−(−11)]=4(x−(−6))

y+11=4x+6y+11=4x+6

⇒4x−y=11−6⇒4x−y=11−6

⇒4x−y=5⇒4x−y=5

Hence 4x−y=54x−y=5 is the required equation.