Question

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) $2x+13=0$

Answer 1

Hint: 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.