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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)` |
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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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