Home
Class 12
MATHS
A tangent to the hyperbola y = (x+9)/(x+...

A tangent to the hyperbola `y = (x+9)/(x+5)` passing through the origin is

A

`x+25y=0`

B

`5x+y=0`

C

`5x-y=0`

D

`x-25y=0`

Text Solution

Verified by Experts

Promotional Banner

Topper's Solved these Questions

  • DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS

    ARIHANT MATHS|Exercise EXAMPLE|6 Videos
  • DY / DX AS A RATE MEASURER AND TANGENTS, NORMALS

    ARIHANT MATHS|Exercise SINGLE OPTION CORRECT TYPE QUESTIONS|10 Videos
  • DIFFERENTIATION

    ARIHANT MATHS|Exercise Exercise For Session 10|4 Videos
  • ELLIPSE

    ARIHANT MATHS|Exercise Exercise (Questions Asked In Previous 13 Years Exam)|26 Videos

Similar Questions

Explore conceptually related problems

To find the point of contact P (x_1, y_1) of a tangent to the graph of y = f(x) passing through origin O, we equate the slope of tangent to y = f(x) at P to the slope of OP. Hence we solve the equation f' (x) = f(x_1)/x_1 to get x_1 and y_1 .Now answer the following questions (7 -9): The equation |lnmx|= px where m is a positive constant has a single root for

A hyperbola passes through the point P(sqrt2,sqrt3) and has foci at (pm2,0) . Then the tangent to this hyperbola at P also passes through the point

The tangent at a point P on the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 passes through the point (0,-b) and the normal at P passes through the point (2asqrt(2),0) . Then the eccentricity of the hyperbola is

Find the equation of the line perpendicular to the x-axis and : passing through the origin.

Find the equation of the straight line perpendicular to y-axis and : passing through the origin.

Find the equation of the plane through the line of intersection of the planes x+2y+3z+4=0 and x-y+z+3=0 and passing through the origin.

The equation of transverse axis of hyperbola (passing through origin) having asymptotes 3x-4y=1 and 4x-3y=6 is ax+by-c=0, a, b in N and gcd(a, b, c)=1 then the value of a+b+c is ___

The co-ordinates of the centre of the smallest circle passing through the origin and having y = x +1 as a diameter are :

Find the equation of a line parallel to x-axis and passing through the origin.

A square of side a lies above the X- axis and has one vertex at the origin . The side passing through the origin makes an angle pi//6 with the positive direction of X-axis .The equation of its diagonal not passing through the origin is y(sqrt(3)-1)-x(1-sqrt(3))=2a y(sqrt(3)+1) +x(1-sqrt(3))=2a y(sqrt(3)+1)+x(1+sqrt(3)) =2a y(sqrt(3)+1)+x(sqrt(3)-1)=2a