If the subnormal at any point on the curve y^ | vedclass

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

Question

(B) Given the equation of the curve is $y^n = ax$. Differentiating both sides with respect to $x$,we get: $n y^{n-1} \frac{dy}{dx} = a$ $\frac{dy}{dx}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Given the equation of the curve is $y^n = ax$. Differentiating both sides with respect to $x$,we get: $n y^{n-1} \frac{dy}{dx} = a$ $\frac{dy}{dx} = \frac{a}{n} y^{1-n}$ The length of the subnormal is defined as $|y \frac{dy}{dx}|$. Substituting the value of $\frac{dy}{dx}$: $	ext{Length of subnormal} = |y \cdot \frac{a}{n} y^{1-n}| = |\frac{a}{n} y^{2-n}|$. For the subnormal to be constant,it must be independent of the variable $y$. This occurs when the exponent of $y$ is $0$,i.e.,$2 - n = 0$. Therefore,$n = 2$. Hence,the correct option is $B$.

If the subnormal at any point on the curve $y^n = ax$ is constant,then the value of $n$ is:

Similar Questions

If the curves $x^2+p y^2=1$ and $q x^2+y^2=1$ are orthogonal to each other,then

The coordinates of a point on the curve $x=a(\theta+\sin \theta), y=a(1-\cos \theta)$ where the tangent is inclined at an angle $\frac{\pi}{4}$ to the positive $X$-axis, are

$P$ is the point of contact of the tangent drawn from the origin to the curve $y = \log_{e} x$. The length of the perpendicular drawn from the origin to the normal at $P$ is

The locus of the point on the curve $y = \sin x$ where the tangent drawn at that point always passes through the point $(0, \pi)$ is

Show that the normal at any point $\theta$ to the curve $x=a \cos \theta+a \theta \sin \theta, y=a \sin \theta-a \theta \cos \theta$ is at a constant distance from the origin.

Vedclass Test Series

Exam Paper Generator

Online Exam Module