The length of the subnormal at any point on t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the curve $y = (\frac{x}{2024})^k$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = k(\frac{x}{2024})^{k-1} \cdot \frac{1}{2024}

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) Given the curve $y = (\frac{x}{2024})^k$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = k(\frac{x}{2024})^{k-1} \cdot \frac{1}{2024} = \frac{k x^{k-1}}{(2024)^k}$. The length of the subnormal is given by the formula $|y \frac{dy}{dx}|$. Substituting the values,we get $|(\frac{x}{2024})^k \cdot \frac{k x^{k-1}}{(2024)^k}| = |\frac{k x^{2k-1}}{(2024)^{2k}}|$. For the length of the subnormal to be constant,it must be independent of $x$. This implies that the exponent of $x$ must be $0$,so $2k - 1 = 0$. Solving for $k$,we get $k = \frac{1}{2}$.

The length of the subnormal at any point on the curve $y = (\frac{x}{2024})^k$ is constant if the value of $k$ is

Similar Questions

The point$(s)$ on the curve $y^3 + 3x^2 = 12y$ where the tangent is vertical (parallel to $y$-axis) is (are):

Consider the curve $y = b e^{-\frac{x}{a}}$, where $a$ and $b$ are non-zero real numbers. Then:

If the normal to the curve $y=f(x)$ at the point $(3,4)$ makes an angle $\left(\frac{3 \pi}{4}\right)^{C}$ with the positive $X$-axis,then $f^{\prime}(3)$ is equal to

If $y=2x$ is a tangent to the curve $y^2=ax^3+b$ at $(1,2)$, then $(a, b)=$

If the tangent to the curve $y = x^3 + ax - b$ at the point $(1, -5)$ is perpendicular to the line $-x + y + 4 = 0$,then which one of the following points lies on the curve?

Vedclass Test Series

Exam Paper Generator

Online Exam Module