Question

For the curve y=4x^(3)-2x^(5) find the equation of tangent which passes through the origin.​

Answer 1

Given , curve is y = 4x³- 2x^5. ..(1)

Let the required point on the given curve be (α,β).

β = 4α³- 2α^5 ..........(2)

Differentiating both sides of (1) w.r.t x, we get

dy/dx = 12x²- 10x^4

(dy/dx)_(α,β)= 12α² - 10α^4

Now, equation of tangent at (α,β) is y - β= (dy/dx)_(α,β) (x - α)

i.. y - β = (12α² - 10β^4) (x - α) ..(3)

Given that tangent passes through the point (0,0).

Putting x = 0 and y = 0 in (3), we get

0 - β= (12α² - 10β^4) (0 - α)

β = 12α³ - 10α^5 ....(4)

From (2) and (4), we get

4α³ - 2α^5 = 12α³ - 10β^5

8α^5 - 8β³ = 0

8α³(α²- 1) = 0

α = 0, 1,-1

Putting α = 0 in (2), we get β = 0.

Putting α = 1 in (2), we get β = 2.

Putting α = -1 in (2), we get β = -2

.(0,0), (1,2) and (-1,-2) are the points on the curve.

Also, all the three points satisfy (4).

=> Hence, required points are (0,0), (1,2) and (-1,-2).