Math, asked by krish78619, 13 days ago

Find the equation of the tangent through a line y = 5x⁴ + 9x⁶ =0 at ( 1,2). Hint : Use derivatives ​

Answers

Answered by sameer6073
0

Answer:

The equation of the given curve is y=x

2

−2x+7.

On differentiating with respect to x, we get:

dx

dy

=2x−2

The equation of the line is 2x−y+9=0.⇒y=2x+9

This is of the form y=mx+c.

Slope of the line =2

If a tangent is parallel to the line 2x−y+9=0, then the slope of the tangent is equal to the slope of the line.

Therefore, we have: 2=2x−2

⇒2x=4⇒x=2

Now, at x=2

⇒y=2

2

−2×2+7=7

Thus, the equation of the tangent passing through (2,7) is given by,

y−7=2(x−2)

⇒y−2x−3=0

Hence, the equation of the tangent line to the given curve (which is parallel to line (2x−y+9=0) is y−2x−3=0.

Step-by-step explanation:

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Answered by shivasinghmohan629
0

Step-by-step explanation:

Differentiate the equation w.r.t. x, we get.

dy/dx = 20x³ +54x5.

Put the value of (x, y) = (1, 2) in the equation, we get.

dy/dx = 20(1)³ + 54(1)5.

dy/dx=

= 20 + 54.

dy/dx = 74.

As we know that, dy/dx is the slope of the equation = m.

→ m = 74.

As we know that,

Formula of equation of tangent.

(y - y₁) = m(x - x₁).

Put the values in the equation, we get.

(y-2) = 74(x - 1).

⇒ y - 2 = 74x - 74.

→ 74x - 74-y + 2 = 0.

Equation of tangent.

(1) = Equation of tangent to the curve y = f(x) at p(x₁, y₁) is (y - y₁) = m(x - x₁).

(2) The tangent at (x₁, y₁) is parallel to x-axis : dy/dx = 0.

(3) = The tangent at (x₁, y₁) is parallel to

y-axis : dy/dx = 0.

(4) = The tangent line makes equal angles with the axis : dy/dx = + 1.

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