1.

if `y=mx+7sqrt(3)` is normal to `(x^(2))/(18)-(y^(2))/(24)=1` then the value of m can beA. `(3)/(sqrt(5)`B. `(sqrt(15))/(2)`C. `(2)/(sqrt(5))`D. `(sqrt(5))/(2)`

Answer» 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))`


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