Question

$T$ is a point on the tangent to a parabola $y^2=4ax$ at its point $P$. $TL$ and $TN$ are the perpendiculars on the focal radius $SP$ and the directrix of the parabola respectively. Then

• A. $SL=2\left(TN\right)$
• B. $3\left(SL\right)=2\left(TN\right)$
• C. $SL=TN$
• D. $2\left(SL\right)=3\left(TN\right)$

Answer 1

The correct option is C $SL=TN$

$P$ is tangent

$S$ is focus

tangent at $P$

$ty=x+a{t}^2$

Let $T$ be $(x_1,y_1)$ lie on tangent

$t{y}_1=x_1+a{t}^2.....\left(1\right)$

Slope of $Sp=\frac{y_2-y_1}{x_2-x_1}=\frac{2at-0}{a{t}^2-a}=\frac{2t}{t^2-1}$

Slope of $SP$ slope of $TC=-1$

${}^m{C}_p={}^m{T}_2=-1$

$\frac{2t}{t^2-1}={}^m{T}_2=-1$

${}^m{T}_c=\frac{-1}{\frac{2t}{t^2-1}}=\frac{1-t^2}{2t}$

equation of $TC\Rightarrow y-y_1=m(x-x_1)$

$y-y_1=\frac{1-t^2}{2t}=\frac{1-t^2}{2t}$

equation of $TC\Rightarrow y-y_1=m(x-x_1)$

$y-y_1=\frac{1-t^2}{2t}(x-x_1)$

$2yt-2y_{1}t=1-t^2(x-x_1)$

$2yt-2y_{1}t+x(t^2-1)-x_1(t^2-1)=0$

$SL$ is perpendicular to $TL$ $LN$ is perpendicular to $TN$

$d_1=SL=\left|\frac{-2t{y}_1+a(t^2-1)-x_1(t^2-1)}{\sqrt{4t^2+(t^2-1{)}^2}}\right|$ $d_2=TN=\left|\frac{x_1+a}{\sqrt{1^2+a^2}}\right|$

$=\left|\frac{-2(x_1+at{)}^2+a(t^2-1)-x_1{t}^2+x_1}{\sqrt{4t^2+t^4+1-2t^2}}\right|$

$=\left|\frac{-2x_1-2a{t}^2+a{t}^2-a-x_1{t}^2+x_1}{\sqrt{t^4+2t^2+1}}\right|$ $\therefore SL=TN$

$=\left|\frac{-a{t}^2-x_1-a-x_1{t}^2}{\sqrt{(t^2+1{)}^2}}\right|=\left|\frac{-(x_1+a)(1+t^2)}{(1+t^2)}\right|=x_1+a$