Home
Class 12
MATHS
if y=mx+7sqrt(3) is normal to (x^(2))/(1...

if `y=mx+7sqrt(3)` is normal to `(x^(2))/(18)-(y^(2))/(24)=1` then the value of m can be

A

`(3)/(sqrt(5)`

B

`(sqrt(15))/(2)`

C

`(2)/(sqrt(5))`

D

`(sqrt(5))/(2)`

Text Solution

Verified by Experts

Given equation of hyperbola, is `(x^(2))/(24)-(y^(2))/(18)=1 " …(i)" `
Since, the equation of the normals of slope m to the hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1`, are given by `y=mxpm(m(a^(2)+b^(2)))/(sqrt(a^(2)-b^(2)m^(2)))`
`therefore ` Equation of normals of slope m, to the hyperbola (i), are
`y=mx pm (m(24+18))/(sqrt(24-m^(2)(18)))" ...(ii)" `
`because " Line " y=mx+7sqrt(3)` is normal to hyperbola (i)
` therefore ` On comparing with Eq. (ii), we get
` pm (m(42))/(sqrt(24-18m^(2)))=7sqrt(3)`
`rArr pm(6m)/(sqrt(24-18m^(2)))=sqrt(3)`
`rArr (36m^(2))/(24-18m^(2))=sqrt(3) ` [squaring both sides]
`rArr 12m^(2)=24-18m^(2)`
`rArr 30m^(2)=24`
`rArr 5m^(2)=4 rArr m=pm (2)/(sqrt(5))`
Promotional Banner

Similar Questions

Explore conceptually related problems

The line y=m x-((a^2-b^2)m)/(sqrt(a^2+b^2m^2)) is normal to the ellise (x^2)/(a^2)+(y^2)/(b^2)=1 for all values of m belonging to (a) (0,1) (b) (0,oo) (c) R (d) none of these

If the straight line 4ax+3by=24 is a normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 (agtb) , then find the the coordinates of focii

If (x+1, y-2)=(3,1), find the values of x and y.

Find the normal to the ellipse (x^2)/(18)+(y^2)/8=1 at point (3, 2).

If the tangent at any point (4m^2,8m^3) of x^3-y^2=0 is a normal to the curve x^3-y^2=0 , then find the value of mdot

If a normal to the ellipse (x^(2))/(a^(2))+(y^(2))/(b^(2))=1 is 4-3y=7 and its ecentricity is (sqrt(7))/(4) , then the volume of L.R can be

If x+4y=14 is a normal to the curve y^2=alphax^3-beta at (2,3), then the value of alpha+beta is 9 (b) -5 (c) 7 (d) -7

The values of m for which the lines y = mx + 2 sqrt5 touches the hyperbola 16 x^(2) - 9y^(2) = 144 are the roots of x^(2) - (a+b) x-4 = 0 then the value of (a+b) is