Home
Class 12
MATHS
Two tangents to the hyperbola (x^(2))/(2...

Two tangents to the hyperbola `(x^(2))/(25) -(y^(2))/(9) =1`, having slopes 2 and m where `(m ne 2)` cuts the axes at four concyclic points then the slope m is/are

A

`-(1)/(2)`

B

`-2`

C

`(1)/(2)`

D

2

Text Solution

Verified by Experts

The correct Answer is:
C

Tangents having slope `'2', y = 2x +- sqrt(100-9) = 2x +- sqrt(91)`
It meets the axis at `A -= (+-(sqrt(91))/(2),0),B (0,+-sqrt(91))`
Tangents having slope `'m', y = mx+- sqrt(25 m^(2)-9)`
It meets the axis at `D(+-(sqrt(25m^(2)-9))/(m),0), C -= (0,+-sqrt(25m^(2)-9))`
ABCD is cyclic quadrilateral
`rArr OA xx OD xx OB xx OC`
`rArr (sqrt(91)(sqrt(25m^(2)-9)))/(2m) =sqrt(91) (sqrt(25m^(2)-9))`
`rArr 2m = 1 rArr m = (1)/(2)`
Promotional Banner

Similar Questions

Explore conceptually related problems

Two tangents to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 having m_1a n dm_2 cut the axes at four concyclic points. Fid the value of m_1m_2dot

Find the equation of tangents to hyperbola x^(2)-y^(2)-4x-2y=0 having slope 2.

The eccentricity of the hyperbola (y^(2))/(9)-(x^(2))/(25)=1 is …………………

If it is possible to draw the tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1 having slope 2, then find its range of eccentricity.

Find the equation of normal to the hyperbola 3x^2-y^2=1 having slope 1/3dot

Find the equation of normal to the hyperbola 3x^2-y^2=1 having slope 1/3dot

If lines 2x-3y+6=0 and kx+2y+12=0 cut the coordinate axes in concyclic points, then the value of |k| is

If m is the slope of a tangent to the hyperbola (x^(2))/(a^(2)-b^(2))-(y^(2))/(a^(3)-b^(3)) =1 where a gt b gt 1 when

If y=m x+c is tangent to the hyperbola (x^2)/(a^2)-(y^2)/(b^2)=1, having eccentricity 5, then the least positive integral value of m is_____

Tangents are drawn to the hyperbola x^(2)/9 - y^(2)/4 parallel to the straight line 2x - y= 1 . One of the points of contact of tangents on the hyperbola is