Solveeit Logo
Login

Question

Question: If the relation between subnormal SN and subtangent ST at any point S on the curve by<sup>2</sup> =...

If the relation between subnormal SN and subtangent ST at any point S on the curve

by2 = (x + a)3 is p(SN) = q(ST)2, then p/q =

A

a27b\frac{a}{27b}

B

8a27b\frac{8a}{27b}

C

8b27a\frac{8b}{27a}

D

8b27\frac{8b}{27}

Answer

8b27\frac{8b}{27}

Explanation

Solution

by2 = (x + a)3

differentiating both sides

2by dydx\frac{dy}{dx} = 3(x + a)2.1

dydx\frac{dy}{dx} = 32\frac{3}{2} (x+a)2by\frac{(x + a)^{2}}{by}

length of subnormal = SN = y dydx\frac{dy}{dx}=32\frac{3}{2} (x+a)2b\frac{(x + a)^{2}}{b}

and length of subtangent = ST = ydxdy\frac{dx}{dy}=2by23(x+a)2\frac{2by^{2}}{3(x + a)^{2}}

pq\frac{p}{q} = (ST)2(SN)\frac{(ST)^{2}}{(SN)} (given)

= (2by2)2.2b{3(x+a)2}2.3(x+a)2\frac{(2by^{2})^{2}.2b}{\{ 3(x + a)^{2}\}^{2}.3(x + a)^{2}}

= 8b27\frac{8b}{27} {(x+a)3}2(x+a)6\frac{\{(x + a)^{3}\}^{2}}{(x + a)^{6}} {using by2 = (x + a)3

= 8b27\frac{8b}{27}

pq\frac{p}{q} = 8b27\frac{8b}{27}