If T is the length of the subtangent drawn at | vedclass

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

Question

(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$ Let $P(h, k)$ be a point on the curve. Differentiating equation $(i)$ with respect to

Vedclass · Accepted Answer

$2 N$

Vedclass · Answer

(C) The equation of the given curve is $3 y^2 = 4 x^3$ ...$(i)$ Let $P(h, k)$ be a point on the curve. Differentiating equation $(i)$ with respect to $x$,we get $6 y \frac{dy}{dx} = 12 x^2$,which implies $\frac{dy}{dx} = \frac{2 x^2}{y}$. At point $(h, k)$,the slope $m = \frac{2 h^2}{k}$. The length of the subtangent $T = \left| \frac{k}{m} \right| = \left| \frac{k}{2 h^2 / k} \right| = \frac{k^2}{2 h^2}$. Since $3 k^2 = 4 h^3$,we have $k^2 = \frac{4}{3} h^3$. Substituting this into $T$,we get $T = \frac{4 h^3}{3(2 h^2)} = \frac{2}{3} h$. The length of the subnormal $N = |k m| = \left| k \cdot \frac{2 h^2}{k} \right| = 2 h^2$. We need to find the relation between $T$ and $N$. From $N = 2 h^2$,we have $h^2 = \frac{N}{2}$,so $h = \sqrt{\frac{N}{2}}$. Then $T = \frac{2}{3} \sqrt{\frac{N}{2}}$. Squaring both sides,$T^2 = \frac{4}{9} \cdot \frac{N}{2} = \frac{2 N}{9}$. Thus,$9 T^2 = 2 N$,which can be written as $(3 T)^2 = 2 N$. Comparing this with $(\beta T)^2 = 2 N$,we find $\beta = 3$. Therefore,the correct option is $C$.

If $T$ is the length of the subtangent drawn at any point on the curve $3 y^2 = 4 x^3$ and $N$ is the length of the subnormal at the same point,then $(\beta T)^2 =$

Similar Questions

If $P$ and $Q$ are two different points on the curve $y = x^3 - x$ such that the tangent at $P$ intersects the curve again at $Q$,then $\frac{m_{OQ} + 1}{m_{OP} + 1}$ is equal to,where $O$ is the origin and $m_{AB}$ represents the slope of the line segment $AB$.

$A(1, -3)$ and $B(4, 3)$ are two points on the curve $y = x - \frac{4}{x}$. The points on the curve,the tangents at which are parallel to the chord $AB$,are

The angle of intersection of the curves $y^2 = \frac{2x}{\pi}$ and $y = \sin x$ is

Find the slope of the tangents to the curve $y = (x + 1)(x - 3)$ at the points where it meets the $x$-axis.

If the two curves $x=y^2$ and $xy=k$ cut each other at right angles, then a possible value of $8k^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module