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If a, b and c are fixed real numbers and l and \mathrm{m} are variable real numbers satisfying \mathrm{a m l+c l^2-b m^2+l=0}, then variable straight line \mathrm{l x+m y+1=0 }always touches a fixed parabola, whose axis is parallel to:

Option: 1

y-axis


Option: 2

 x-axis
 


Option: 3

 y=x
 


Option: 4

 y=-x


Answers (1)

best_answer

Any parabola whose axis is parallel to \mathrm{x}-axis, will be of the form \mathrm{y^2+A y+B x+C=0 }, put \mathrm{x=\frac{-(m y+1)}{l} } in the equation of parabola to get
\mathrm{y^2+A y-\frac{B}{l}(m y+1)+C=0 \text { or } l^2+(A l-B m) y+C l-B=0 }If the given line touches the parabola, then the above equation should have real and equal roots i.e.,
\mathrm{\begin{aligned} & (A l-B m)^2-4 l(C l-B)=0 \\ & \left(A^2-4 C\right) l^2-2 A B l m+4 B l+B^2 m^2=0 \end{aligned} }
or
This is the condition of tangency. Comparing it with the given condition on a, b, c, l and \mathrm{\mathrm{m} } we get\mathrm{ A^2-4 C=c k,-2 A B=a k, 4 B=k, B^2=-b k, }where k is proportionality constant. Solving all these, we get k=0 or k=-16 b.
k=0 does not give a parabola, so using k=-16 b
we get the parabola \mathrm{(y-a)^2=4 b(x-c). }
Which is a fixed parabola whose axis is parallel to \mathrm{\mathrm{x}-axis }(as a, b and c are fixed).
 

Posted by

Deependra Verma

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