Given: Equation of curve is
x2+3y=3 and the tangent to the curve
x2+3y=3 is parallel to the line y - 4x + 5 = 0.
The given line y - 4x + 5 = 0 can be re-written as:
⇒ y = 4x - 5
Now by comparing the above equation of line with y = mx + c we get,
⇒ m = 4 and c = 5
∵ The line y - 4x + 5 = 0 is parallel to the tangent to the curve
x2+3y=3 As we know that if two lines are parallel then their slope is same.
So, the slope of the tangent to the curve
x2+3y=3 is m = 4.
Let the point of contact be
(x1,y1) As we know that slope of the tangent at any point say
(x1,y1) to a curve is given by:
m=[](x1,‌‌y1) So by differentiating the equation
x2+3y=3 with respect to x we get
⇒2x+3⋅=0 ⇒=‌− ⇒[](x1,y1)=‌− ∵ Slope of tangent to the curve x + 3y = 3 is m = 4
⇒4=‌− ⇒x1=−6 Now by substituting
x1=−6 int equation
x2+3y=3 we get
⇒36+3y=3 ⇒y=−11 So, the point of contact is
(−6,−11) As we know that equation of tangent at any point say
(x1,y1) is given by:
y−y1=[](x1,‌y1)⋅(x−x1) ⇒y+11=4⋅(x+6) ⇒4x−y+13=0 Hence, the equation of the required tangent is
4x−y+13=0