A tangent to the curve y = f(x) at P(x, y) me | vedclass

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

Question

(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \rig

Vedclass · Accepted Answer

$\left( 2, \frac{1}{8} \right)$

Vedclass · Answer

(C) Let the tangent at $P(x, y)$ be $Y - y = f'(x)(X - x)$.For $A$ (where $Y=0$),$X = x - \frac{y}{f'(x)}$. So $A = \left( x - \frac{y}{f'(x)}, 0 \right)$.For $B$ (where $X=0$),$Y = y - x f'(x)$. So $B = (0, y - x f'(x))$.Given $AP : BP = 1 : 3$,by section formula for $P(x, y)$ dividing $AB$ in ratio $1:3$:$x = \frac{1 \cdot 0 + 3 \cdot (x - y/f'(x))}{1 + 3} \implies 4x = 3x - \frac{3y}{f'(x)} \implies x = -\frac{3y}{f'(x)}$.Thus,$\frac{dy}{dx} = -\frac{3y}{x}$.Separating variables: $\int \frac{dy}{y} = -3 \int \frac{dx}{x} \implies \ln|y| = -3 \ln|x| + C \implies y = \frac{k}{x^3}$.Given $f(1) = 1$,we get $1 = \frac{k}{1^3} \implies k = 1$. So $y = \frac{1}{x^3}$.Checking the options,for $x=2$,$y = \frac{1}{2^3} = \frac{1}{8}$.Thus,the curve passes through $\left( 2, \frac{1}{8} \right)$.

$A$ tangent to the curve $y = f(x)$ at $P(x, y)$ meets the $x-$axis at $A$ and the $y-$axis at $B$. If $AP : BP = 1 : 3$ and $f(1) = 1$,then the curve also passes through the point:

Similar Questions

Number of values of $m \in N$ for which $y = e^{mx}$ is a solution of the differential equation $D^3y - 3D^2y - 4Dy + 12y = 0$ is

The solution of the differential equation $x \frac{d^2y}{dx^2} = 1$,given that $y = 1$ and $\frac{dy}{dx} = 0$ when $x = 1$,is

The equation to the orthogonal trajectories of the system of parabolas $y = ax^2$ is

The rate of growth of bacteria is proportional to the number present. If initially there were $1000$ bacteria and the number doubles in $1$ hour,then the number of bacteria after $2 \frac{1}{2}$ hours is (Given $\sqrt{2} = 1.414$):

If the function $y = e^{4x} + 2e^{-x}$ is a solution of the differential equation $\frac{\frac{d^3y}{dx^3} - 13\frac{dy}{dx}}{y} = K$,then the value of $K$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module