Question

find the equation to the tangent of the curve y^2=16x at (4,8)​

Answer 1

Answer:

x−y+4=0

Tangent to the parabola y

2

=16x is y=mx+

m

4

solve it by curve xy=−4

i.e. mx

2

+

m

4

x+4=0

condition of common tangent is D=0

∴m

3

=1

⇒m=1

∴ equation of common tangent is y=x+4.

Answer 2

Answer:

x+y = 4

Step-by-step explanation:

The given curve is y =f(x) with factor A (x1, y1). Therefore, the slope of the tangent becomes (dy/dx)x = x1 ; y = y1.  

Now when you consider that we've observed a way to calculate the slope, we are able to truely pass ahead with following the idea of coordinate geometry in keeping with which an equation of a line is

(y − y1) = m(x − x1) , wherein m is the slope.

We can now replace “

m” from the equation above to the (dy/dx)x = x1 ; y = y1 . Now, the brand new equation of the tangent to the curve at factor A (x1, y1) is

(y − y1) = (dy/dx)x = x1 ; y = y1 (x − x1).

The above-stated equation is the equation of the tangent formula. In summary, comply with those 3 easy steps to locate the equation of the tangent to the curve at factor A (x1, y1).

y² = 16x

2y $\frac{dy}{dx}$ =16

$\frac{dy}{dx}$ =18

$\frac{dy}{dx}$  4,8 = 8/8 =1

slope of normal = -$\frac{dy}{dx}$

  that equal to -1

slope of normal is -1

point (4,8)

equation of normal

y-8 = -1(x-4)

y-8=-x+4

x+y = 4

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