A tangent to the curve y=f(x) cuts the line y=x at a point which is at a distance of 1 unit from y-axis. The equation of the curve is

  • Option 1)

    \frac{x-1}{y-1}=c

  • Option 2)

    \frac{x}{y}=c

  • Option 3)

    xy=c

  • Option 4)

    None

 

Answers (1)

As we discussed in concept

Solution of Differential Equation -

\frac{\mathrm{d}y }{\mathrm{d} x} =f\left ( ax+by+c \right )

put

 Z =ax+by+c

 

 

- wherein

Equation with convert to

\int \frac{dz}{bf\left ( z \right )+a} =x+c

 

 Let the point (1,1) (because line is y=x)

\therefore equation of tangent

(y-1)=\frac{dy}{dx}(x-1)                where \frac{dy}{dx}\:=\:f^{'}(x)

                                                         = slope of tangent.

\frac{dy}{dx}=\frac{y-1}{x-1}

\therefore\:\int \frac{dy}{y-1}\:=\int \frac{dx}{x-1}

log(y-1)=log(x-1)+c\:\:\:\:\:\:\:\:\:\:\:\:c=log\:\frac{y-1}{x-1}

 

 

 


Option 1)

\frac{x-1}{y-1}=c

This option is correct.

Option 2)

\frac{x}{y}=c

This option is incorrect.

Option 3)

xy=c

This option is incorrect.

Option 4)

None

This option is incorrect.

Preparation Products

Knockout BITSAT 2020

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 1999/-
Buy Now
Knockout BITSAT 2021

It is an exhaustive preparation module made exclusively for cracking BITSAT..

₹ 4999/- ₹ 2999/-
Buy Now
Knockout BITSAT-JEE Main 2021

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 27999/- ₹ 16999/-
Buy Now
Knockout BITSAT-JEE Main 2020

An exhaustive E-learning program for the complete preparation of JEE Main and Bitsat.

₹ 14999/- ₹ 7999/-
Buy Now
Exams
Articles
Questions