Home
Class 12
MATHS
lf length of tangent at any point on th ...

lf length of tangent at any point on th curve `y=f(x)` intercepted between the point and the x-axis is of length 1. Find the equation of the curve.

Text Solution

Verified by Experts

The correct Answer is:
`(sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))= pm x+c)`
Since, the length of tangent = `abs(ysqrt(1+(dx/dy)^(2)))=1`
`rArr y^(2) (1+((dx)/dy)^2)=1`
`therefore dy/dx=pm y/(sqrt(1-y^(2)))`
`rArr int (sqrt(1-y^(2)))/y dy =pm int xdx `
`rArr int (sqrt(1-y^(2)))/y dy =pm x+C`
Lut `y=sin theta rArr dy = cos theta d theta`
`therefore int (cos theta)/(sin theta)cdot cos theta d theta = pm x+C`
`therefore int (cos^(2) theta)/(sin^(2) theta)cdot sin theta d theta = pm x+C`
Again put `cos theta =t rArr -sin theta d theta =dt`
`therefore -int t^(2) /(1-t^(2))dt=pm x+C`
`rArr int (1-1/(1-t^(2))) dt =pm x+C`
`rArr t-log abs((1=t)/(1-t))=pmx+C`
`rArr sqrt(1-y^(2))-log abs((1+sqrt(1-y^(2)))/(1-sqrt(1-y^(2))))=pm x+C`
Promotional Banner

Topper's Solved these Questions

  • DIFFERENTIAL EQUATIONS

    IIT JEE PREVIOUS YEAR|Exercise TOPIC 3 APPLICATIONS OF HOMOGENEOUS DIFFERENTIAL EQUATIONS <br> FILL IN THE BLANK|1 Videos

  • DEFINITE INTEGRATION

    IIT JEE PREVIOUS YEAR|Exercise LIMITS AS THE SUM|6 Videos

  • ELLIPSE

    IIT JEE PREVIOUS YEAR|Exercise EQUATION OF CHORD OF CONTACT, CHORD BISECTED AT A GIVEN POINT & DIAMETER|3 Videos

Similar Questions

Explore conceptually related problems

If length of tangent at any point on th curve y=f(x) intercepted between the point and the x -axis is of length 1. Find the equation of the curve.

If length of tangent at any point on the curve y=f(x) intercepted between the point and the x -axis is of length l. Find the equation of the curve.

The length of the subtangent at any point on the curve y=be^(x/3) is

The length of the subtangent at any point (x_1,y_1) on the curve y=5^x is

A curve y=f(x) passes through the point P(1,1) . The normal to the curve at P is a(y-1)+(x-1)=0 . If the slope of the tangent at any point on the curve is proportional to the ordinate of the point. Determine the equation of the curve. Also obtain the area bounded by the y-axis, the curve and the normal to the curve at P .

The slope of tangent at a point P(x, y) on a curve is - x/y . If the curve passes through the point (3, -4) , find the equation of the curve.

A curve y=f(x) passes through point P(1,1). The normal to the curve at P is a (y-1)+(x-1)=0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point,then the equation of the curve is

A normal at any point (x,y) to the curve y=f(x) cuts a triangle of unit area with the axis,the differential equation of the curve is

The length of normal at any point to the curve,y=c cosh((x)/(c)) is

The length of tangent at a point P(x_(1),y_(1)) to the curve y=f(x) ,having slope m at P is