Find all points on the curve y = 4x^3 - 2x^5 | vedclass

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(A) Let $P(h, k)$ be a point on the curve $y = 4x^3 - 2x^5$. Since $P$ lies on the curve,we have $k = 4h^3 - 2h^5$ ..... $(1)$. The slope of the tange

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$(0, 0), (1, 2), (-1, -2)$

Vedclass · Answer

(A) Let $P(h, k)$ be a point on the curve $y = 4x^3 - 2x^5$. Since $P$ lies on the curve,we have $k = 4h^3 - 2h^5$ ..... $(1)$. The slope of the tangent at $P(h, k)$ is given by $\frac{dy}{dx} = 12x^2 - 10x^4$. At $x = h$,the slope $m = 12h^2 - 10h^4$. The equation of the tangent at $P(h, k)$ is $y - k = (12h^2 - 10h^4)(x - h)$. Since the tangent passes through the origin $(0, 0)$,we substitute $x = 0$ and $y = 0$: $0 - k = (12h^2 - 10h^4)(0 - h)$ $-k = -12h^3 + 10h^5$ $k = 12h^3 - 10h^5$ ..... $(2)$. Equating $(1)$ and $(2)$: $4h^3 - 2h^5 = 12h^3 - 10h^5$ $8h^5 - 8h^3 = 0$ $8h^3(h^2 - 1) = 0$ This gives $h = 0, h = 1, h = -1$. For $h = 0$,$k = 4(0)^3 - 2(0)^5 = 0$. Point is $(0, 0)$. For $h = 1$,$k = 4(1)^3 - 2(1)^5 = 4 - 2 = 2$. Point is $(1, 2)$. For $h = -1$,$k = 4(-1)^3 - 2(-1)^5 = -4 + 2 = -2$. Point is $(-1, -2)$. Thus,the points are $(0, 0), (1, 2), (-1, -2)$.

Find all points on the curve $y = 4x^3 - 2x^5$ at which the tangent passes through the origin.

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