If the relation between the subnormal SN and | vedclass

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

Question

(A) Given the curve equation $by^2 = (x + a)^3$. Differentiating with respect to $x$: $2by \frac{dy}{dx} = 3(x + a)^2$ $\frac{dy}{dx} = \frac{3(x + a)

Vedclass · Accepted Answer

$8b/27$

Vedclass · Answer

(A) Given the curve equation $by^2 = (x + a)^3$. Differentiating with respect to $x$: $2by \frac{dy}{dx} = 3(x + a)^2$ $\frac{dy}{dx} = \frac{3(x + a)^2}{2by}$ The length of the subnormal $SN$ is given by $|y \frac{dy}{dx}|$: $SN = y \cdot \frac{3(x + a)^2}{2by} = \frac{3(x + a)^2}{2b}$ The length of the subtangent $ST$ is given by $|y / \frac{dy}{dx}|$: $ST = y \cdot \frac{2by}{3(x + a)^2} = \frac{2by^2}{3(x + a)^2}$ Since $by^2 = (x + a)^3$,we substitute this into the expression for $ST$: $ST = \frac{2(x + a)^3}{3(x + a)^2} = \frac{2(x + a)}{3}$ We are given the relation $p(SN) = q(ST)^2$,which implies $\frac{p}{q} = \frac{(ST)^2}{SN}$. Substituting the expressions for $SN$ and $ST$: $\frac{p}{q} = \frac{(\frac{2(x + a)}{3})^2}{\frac{3(x + a)^2}{2b}} = \frac{\frac{4(x + a)^2}{9}}{\frac{3(x + a)^2}{2b}} = \frac{4}{9} \cdot \frac{2b}{3} = \frac{8b}{27}$.

If the relation between the subnormal $SN$ and the subtangent $ST$ at any point $S$ on the curve $by^2 = (x + a)^3$ is $p(SN) = q(ST)^2$,then what is the value of $p/q$?

Similar Questions

The equation of the tangent to the curve $y = 3x^3 - 3x^2 + x$ at $x = 1$ is

For the curve ${y^n} = {a^{n - 1}}x$,the subnormal at any point is constant. The value of $n$ must be

If the length of the subtangent at any point to the curve $x y^{n}=a$ is proportional to the abscissa,then $n$ is

If the equation of a tangent drawn to the curve $y=\cos (x+y), -1 \leq x \leq 1+\pi$ is $x+2y=k$, then $k=$

The ratio of the length of the subnormal to the square of the length of the subtangent at any point $P$ on the curve $y^2=(2x+1)^3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module