Question

Let the line $lx+my=1$ cut the parabola ${{y}^{2}}=4ax$ in the points A and B. Normals at A and B meets the point C. Normal from C other than these two meet the parabola at D then the coordinate of D are –
a.$\left( a,2a \right)$
b.$\left( \dfrac{4am}{{{l}^{2}}},\dfrac{4a}{l} \right)$
c.$\left( \dfrac{2a{{m}^{2}}}{{{l}^{2}}},\dfrac{2a}{l} \right)$
d.$\left( \dfrac{4a{{m}^{2}}}{{{l}^{2}}},\dfrac{4am}{l} \right)$

Answer 1

Hint: From the equation $lx+my=1$ get equation for y and put the value of y in ${{y}^{2}}=4ax$ . Simplify and solve the quadratic expression formed. The 3 normals that can be drawn from a point is given by an equation. Simplify the value of y we got according. Thus, find the coordinate of D.

Complete step-by-step answer:
We have been given the equation of the line as $lx+my=1$ .
We have been given the equation of parabola as ${{y}^{2}}=4ax$ .
From the given equation of line, we can write,
$lx+my=1\Rightarrow my=1-lx$
$\therefore y=\dfrac{1-lx}{m}$…………………….. (1)
Now, put this in value of y in ${{y}^{2}}=4ax$.
${{\left( \dfrac{1-lx}{m} \right)}^{2}}=4ax\Rightarrow \dfrac{{{\left( 1-lx \right)}^{2}}}{{{m}^{2}}}=4ax$ .
We know ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2{ab}$
$\begin{aligned} & \therefore 1-2lx+{{l}^{2}}{{x}^{2}}=4a{{m}^{2}} \\\ & {{l}^{2}}{{x}^{2}}-2lx-4{{m}^{2}}xa+1=0 \\\ & \Rightarrow {{x}^{2}}{{l}^{2}}-\left( 2l+4{{m}^{2}}a \right)x+1=0 \\\ \end{aligned}$
Now the above equation is of the form $a{{x}^{2}}+bx+c=0$ . Now by comparing both equation, we get –
$a={{l}^{2}},b=-\left( 2l+4{{m}^{2}}a \right),c=1$ .
Put these values in the quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ .
$\begin{aligned} & x=\dfrac{\left( 2l+4{{m}^{2}}a \right)\pm \sqrt{{{\left( 2l+4{{m}^{2}}a \right)}^{2}}-4\times {{l}^{2}}\times 1}}{2{{l}^{2}}} \\\ & =\dfrac{\left( 2l+4{{m}^{2}}a \right)\pm \sqrt{4{{l}^{2}}+16l{{m}^{2}}a+16{{m}^{4}}{{a}^{2}}-4{{l}^{2}}}}{2{{l}^{2}}} \\\ & =\dfrac{\left( 2l+4{{m}^{2}}a \right)\pm \sqrt{16{{m}^{2}}{{a}^{2}}\left( l+{{m}^{2}} \right)}}{2{{l}^{2}}} \\\ & =\dfrac{\left( 2l+4{{m}^{2}}a \right)\pm 4ma\sqrt{l+{{m}^{2}}}}{2{{l}^{2}}} \\\ \end{aligned}$ $\therefore {{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}$
Thus we got $x=\dfrac{2l+4{{m}^{2}}a\pm 4ma\sqrt{l+{{m}^{2}}}}{2{{l}^{2}}}=\dfrac{l+2{{m}^{2}}a\pm 2ma\sqrt{l+{{m}^{2}}}}{{{l}^{2}}}$
Hence, we get the value of x in the equation of y.
$\begin{aligned} & y=\dfrac{1-lx}{m}=\dfrac{1}{m}\left[ 1-l\left( \dfrac{l+2{{m}^{2}}a\pm 2ma\sqrt{l+{{m}^{2}}}}{{{l}^{2}}} \right) \right] \\\ & y=\dfrac{1}{m}\left[ \dfrac{2{{m}^{2}}a\pm 2ma\sqrt{l+{{m}^{2}}}}{l} \right]=\dfrac{m}{m}\left[ \dfrac{2ma\pm 2a\sqrt{l+{{m}^{2}}}}{l} \right] \\\ \end{aligned}$
$\therefore y=\dfrac{2ma\pm 2a\sqrt{l+{{m}^{2}}}}{l}$……………………… (2)
The 3rd coordinate given to us is D. Let us take the coordinates of D as $\left( {{x}_{3}},{{y}_{3}} \right)$ . Through point C. 3 normals pass.
The equation of normal for the general equation of parabola ${{y}^{2}}=4ax$ is given as,
$y-{{y}_{1}}=\dfrac{-{{y}_{1}}}{2a}\left( x-{{x}_{1}} \right)$ .
Now let us find the equation of normal for point $P\left( at_{1}^{2},2a{{t}_{1}} \right)$. Thus, put $x=at_{1}^{2}$ and ${{y}_{1}}=2a{{t}_{1}}$ in the above equation.
$y-2a{{t}_{1}}=\dfrac{-2a{{t}_{1}}}{2a}\left( x-at_{1}^{2} \right)$
Now cancel out the like terms and simplify the above expression,
$\begin{aligned} & y-2a{{t}_{1}}=-{{t}_{1}}\left( x-at_{1}^{2} \right) \\\ & y-2a{{t}_{1}}=-x{{t}_{1}}+at_{1}^{3} \\\ & \Rightarrow y=-{{t}_{1}}x+at_{1}^{3} \\\ \end{aligned}$
Thus from a point 3 normals can be given by the equation. $y=tx-2at-a{{t}^{3}}$ .
The sum of all values of ${{t}_{1}}$ and t=0.
We know the coordinates of y as $2a{{t}_{1}}$ .
Thus & y=0, Now put & y=0 in equation (2). Thus we can write ${{y}_{1}}+{{y}_{2}}+{{y}_{3}}=0$ .
$\begin{aligned} & {{y}_{1}}=\dfrac{2ma+2a\sqrt{l+{{m}^{2}}}}{l} \\\ & {{y}_{2}}=\dfrac{2ma-2a\sqrt{l+{{m}^{2}}}}{l} \\\ & \therefore \dfrac{2ma+2a\sqrt{l+{{m}^{2}}}}{l}+\dfrac{2ma-2a\sqrt{l+{{m}^{2}}}}{l}+{{y}_{3}}=0 \\\ \end{aligned}$
$\begin{aligned} & \dfrac{2ma+2a\sqrt{l+{{m}^{2}}}+2ma-2a\sqrt{l+{{m}^{2}}}}{l}+{{y}_{3}}=0 \\\ & \dfrac{4ma}{l}={{y}_{3}} \\\ \end{aligned}$
Hence we got the value of ${{y}_{3}}=\dfrac{4ma}{l}$.
Thus from the options we can say that value of ${{x}_{3}}=\dfrac{4a{{m}^{2}}}{{{l}^{2}}}$ .
$\therefore$ The coordinates of D=$\left( \dfrac{4a{{m}^{2}}}{{{l}^{2}}},\dfrac{4am}{l} \right)$.
Therefore, option (d) is the correct answer.

Note: For a given parabola and for a given point let say $\left( b,c \right)$ , the cubic m has 3 roots say ${{m}_{1}},{{m}_{2}},{{m}_{3}}$ . The 3 normals that can be drawn to the parabola have slope ${{m}_{1}},{{m}_{2}},{{m}_{3}}$ . For this cubic, we have ${{m}_{1}},{{m}_{2}},{{m}_{3}}=0$ .