If l, m are variable real numbers such that \[5{{l}^{2}}+6{{m}^{2}}-4lm+3l=0\], the variable line \[lx+my=1\] always touches a fixed parabola, whose axis is parallel to X-axis . The equation of directrix of parabola is
(a) \[6x+7=0\]
(b) \[4x+11=0\]
(c) \[3x+11=0\]
(d) \[2x+13=0\]
VerifiedHint: The rough diagram of given data is
We solve this problem by assuming the equation of parabola whose axis is parallel to X – axis is given as \[{{\left( y-a \right)}^{2}}=4b\left( x-c \right)\] then we use the condition that \[lx+my=1\] is the line equation which touches the parabola which means the tangent. We use the condition of tangent that is if
\[y=mx+c\] is the tangent to \[{{y}^{2}}=4ax\] then
\[\Rightarrow c=\dfrac{a}{m}\]
By using the above condition we find the values of \[a,b,c\] so that the equation of directrix is given as
\[\Rightarrow \left( x-c \right)+b=0\]
Complete step-by-step solution:
Let us assume that the equation of parabola whose axis is parallel to X – axis as
\[\Rightarrow {{\left( y-a \right)}^{2}}=4b\left( x-c \right)\]
Let us modify the above equation as
\[\Rightarrow {{Y}^{2}}=4bX........equation(i)\]
We are given that the line \[lx+my=1\] always touches the parabola.
Let us modify the equation of tangent given in the form of \[X,Y\] as follows
$ \Rightarrow lx+my=1 $
$ \Rightarrow \left( y-a \right)=\dfrac{-l}{m}\left( x-c \right)+\left( \dfrac{1}{m}-a+-\dfrac{lc}{m} \right) $
$ \Rightarrow Y=\dfrac{-l}{m}\left( X \right)+\left( \dfrac{1-am-lc}{m} \right).........equation(ii) $
We know that the condition of tangent that is if \[y=mx+c\] is the tangent to \[{{y}^{2}}=4ax\] then
\[\Rightarrow c=\dfrac{a}{m}\]
By using the above condition to equation (i) and equation (ii) we get
$ \Rightarrow \dfrac{1-am-lc}{m}=\dfrac{b}{\left( \dfrac{-l}{m} \right)} $
$ \Rightarrow -b{{m}^{2}}=l-aml-c{{l}^{2}} $
$ \Rightarrow c{{l}^{2}}-b{{m}^{2}}+aml-l=0..........equation(iii) $
We are given that
\[5{{l}^{2}}+6{{m}^{2}}-4lm+3l=0\]
By comparing the above equation with equation (iii) we get
\[\Rightarrow \dfrac{c}{5}=\dfrac{-b}{6}=\dfrac{a}{-4}=\dfrac{-1}{3}\]
Now, by dividing the terms of each variable we get
$ \Rightarrow c=\dfrac{-5}{3} $
$ \Rightarrow b=2 $
$ \Rightarrow a=\dfrac{4}{3} $
By substituting the above values in the equation (i) we get
\[\Rightarrow {{\left( y-\dfrac{4}{3} \right)}^{2}}=8\left( x+\dfrac{5}{3} \right)\]
We know that the equation of directrix of parabola \[{{\left( y-a \right)}^{2}}=4b\left( x-c \right)\] is given as
\[\Rightarrow \left( x-c \right)+b=0\]
By using the above condition the equation of directrix is given as
$ \Rightarrow x+\dfrac{5}{3}+2=0 $
$ \Rightarrow 3x+11=0 $
Therefore the equation of directrix of given parabola is \[3x+11=0\]
So, option (c) is the correct answer.
Note: Students may make mistakes in taking the condition for tangent as wrong. We have the condition of tangent as if \[y=mx+c\] is the tangent to \[{{y}^{2}}=4ax\] then
\[\Rightarrow c=\dfrac{a}{m}\]
But the equation of parabola we have is
\[\Rightarrow {{\left( y-a \right)}^{2}}=4b\left( x-c \right)\]
So, to apply the condition the equation of tangent should also be in the form
\[\left( y-a \right)=m\left( x-c \right)+C\]
Students may miss this point which results in a wrong answer.
This point needs to be taken care of.
