Find the equations of the tangents to the curve x = 9t2 + 6, y = 6t3 + 3 that pass through the point (15, 9).

The equation of the tangent is .

The given functions are and .

Differentiate the given function with respect to as-

Similarly,

So,

Now, the point occurs on the curve and corresponds to

Hence, the slope of the line tangent to the curve at is 1.

And, the equation of the tangent is .

Learn more about tangents here:

link

Tangents to the curve have slope . We have

The point (15, 9) occurs on the curve and corresponds to

so that the slope of the line tangent to the curve at (15, 9) is 1. Then the equation of this tangent line is

step-by-step explanation:

you divide the denominator by the numerator on both sides and if the quotients are equal then the fractions are equivalent