Question

The condition that the line $lx+my+n=0$ may be a normal to the parabola $y^2=4ax$ is

• A. $a{l}^3+2al{m}^2+m^{2}n=0$
• B. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$
• C. $\frac{a^2}{l^2}+\frac{b^2}{m^2}=\frac{(a^2-b^2{)}^2}{n^2}$
• D. $\frac{a^2}{l^2}-\frac{b^2}{m^2}=\frac{(a^2+b^2{)}^2}{n^2}$

Answer 1

The correct option is A $a{l}^3+2al{m}^2+m^{2}n=0$

The equation of the given parabola is $y^2=4ax$.

The equation of the normal to the given parabola is $y+tx=2at+a{t}^3$, i.e. $tx+y-(2at+a{t}^3)=0$ ....(1)

The equation (1) and the given equation of the normal $lx+my+n=0$ ....(2) represents the same line.

Therefore, $\frac{t}{l}=\frac{1}{m}=\frac{-(2at+a{t}^3)}{n}$

$\Rightarrow \frac{t}{1}=\frac{1}{m}$

$\Rightarrow t=\frac{1}{m}$ ....(3)

and $\frac{1}{m}=\frac{-(2at+a{t}^3)}{n}$

$\Rightarrow -\frac{n}{m}=2at+a{t}^3$

$\Rightarrow -\frac{n}{m}=2a.\left(\frac{l}{m}\right)+a.{\left(\frac{l}{m}\right)}^3$

$\Rightarrow -\frac{n}{m}=\frac{2al}{m}+\frac{a{l}^3}{m^3}$

$\Rightarrow -n{m}^2=2al{m}^2+a{l}^3$

$\Rightarrow a{l}^3+2al{m}^2+n{m}^2=0$