If y=mx+c is tangent to the hyperbola x2a2−y2b2=1having eccentricity 5, then the least positive integral value of m is

If y=mx+c is tangent to the hyperbola x2a2y2b2=1having eccentricity 5, then the least positive integral value of m is

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    Solution:

    e2=b2a2+1b2a2=24

    Condition of tangency is  c2=a2m2b20m2b2a2

    m26

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