Question

Find the equation of the tangents drawn from the point (1, 2) to the curve
y2 - 2x3 - 4y + 8 = 0.​

Answer 1

Answer:

Step-by-step explanation:

’ll assume you mean: y^2 - 2x^3 - 4y + 8 = 0

Differentiate implicitly:

2′−62−4′=0

(2−4)′=62

′=32−2

Let  (,)  be a point on the curve  2−23−4+8=0

Let  ′=  Slope of line through  (,)  and  (1,2) :

32−2=−2−1

(−2)2=32(−1)

Plug into equation of curve and solve for   :

2−23−4+8=0

(−2)2−23+4=0

32(−1)−23+4=0

3−32+4=0

(+1)(−2)2=0

=−1,2

Calculate   :

=−1⇒(−2)2=3(−1)2(−1–1)=−6

=2⇒(−2)2=3(2)2(2–1)=12⇒=2±23‾√

Calculate  ′ :

′=32−2=3(2)2±23‾√=±23‾√

Tangent lines drawn from the point  (1,2) :

=2±23‾√(−1)

                                               hope it helps

Answer 2

Step-by-step explanation:

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