Question

$T$ is a point on the tangent to a parabola $y^2=4ax$ at its point $TL$ 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. $32\left(SL\right)=22\left(TN\right)$
• C. $SL=TN$
• D. $22\left(SL\right)=32\left(TN\right)$

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$\cdot$ Slope of TL$=-1$

$m_{Sp}\cdot m_{TL}=-1$

$\frac{2t}{t^2-1}\cdot m_{TL}=-1$

$m_{TL}=\frac{-1}{\frac{2t}{t^2-1}}=\frac{1-t^2}{2t}$

Equation of $TL\Rightarrow y-y_1=m(x-x_1)$

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

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

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

SL is perpendicular to TL LN is perpendicular to TN

$d_1=SL=\left|\frac{-2+y_1+a(t^2-1)-x_1(t^2-1)}{\sqrt{4t^2+(t^2-1{)}^2}}\right|$ $d_2=TN=\left|\frac{x_1+a}{\sqrt{1^2+0^2}}\right|$

[$\because$ from $1$]

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

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

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

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