Find the equation of the tangent to 3x^2 +4y^2 =7 at P(1, 1).
(pls provide step by step, use implicit differentiation when finding derivative)
First, lets find the implicit derivative of the function. Our equation is:
3x^2 + 4y^2 = 7
Now, lets differentiate both sides with respect to x:
6x + 8yy' = 0
Now, we can find the derivative y' by solving for it:
8yy' = -6x
y' = (-6x)/(8y)
Now, let's substitute the given point P(1, 1) into this expression to find the slope of the tangent at point P:
y'(1) = (-6(1))/(8(1)) = (-6)/8 = -3/4
Now, we have the slope of the tangent at P(1, 1) - it is -3/4. Now, we can use the point-slope form of the equation of a line to find the equation of the tangent. The point-slope form is:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) is our given point. Plugging in the slope and the coordinates of the point, we get:
y - 1 = (-3/4)(x - 1)
Now we can simplify by multiplying both sides by 4:
4(y - 1) = -3(x - 1)
4y-4=-3x+3
Now, move everything to the left side to get it in the form Ax + By = C:
3x + 4y = 7
The equation of the tangent to 3x^2 + 4y^2 = 7 at P(1, 1) is 3x + 4y = 7.