Prove that any hyperbola and its conjugate hyperbola cannot have common normal.
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Consider hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1.` Equation of normal to hyperbola at point `P(a sec theta, b tan theta)` is `ax cos theta+by cot theta=a^(2)+b^(2)" (1)"` Equation of normal to hyperbola `(x^(2))/(a^(2))-(y^(2))/(b^(2))=-1` at point `Q( a tan phi, b sec phi)` is `ax cot phi+"by" cos phi=a^(2)+b^(2)" (2)"` If Eqs. (1) and (2) represent the same straight line, then `(cot phi)/(cos theta)=(cos phi)/(cot theta)=1` `rArr" "tan phi = sec theta and sec phi = tan theta` `rArr" "sec^(2)phi-tan^(2)phi=tan^(2)theta-sec^(2)theta=-1,` which is not possible. Thus, hyperbola and its conjugate hyperbola cannot have common normal.
If a variable line has its intercepts on the coordinate axes e and e^(prime), where e/2a n d e^(prime)/2 are the eccentricities of a hyperbola and its conjugate hyperbola, then the line always touches the circle x^2+y^2=r^2, where r=
Column I|Column II (a)Two intersecting circle| p. have a common tangent (b)Two mutually external circles| q. have a common normal (c)Two circles, one strictly inside the other| r. do not have a common tangent (d)Two branches of a hyperbola| s.do not have a common normal
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