Question

P is point $(a{t}^2,2at)$ on a parabola $y^2=4ax$ and PQ is a chord PT is the tangent at P and QN is the normal at Q if the angle between PT and QN be $\alpha$ and the distance between PT and QN be 'd', then

• A. $0<\alpha <{90}^{\circ }$
• B. $d=\left|\frac{a(1+t^2{)}^{3/2}}{t^2}\right|$
• C. $d=0$
• D. $d=\left|\frac{a(1+t^2{)}^{3/2}}{t}\right|$

Answer 1

The correct option is B $d=\left|\frac{a(1+t^2{)}^{3/2}}{t^2}\right|$

$P(a{t}^2,2at)$ and $Q(\frac{a}{t^2},\frac{-2a}{t})$ as $PQ$ is focal chord parabola $y^2=4ax................\left(1\right)$

Tangent at $P=>y=\frac{x}{t}+at$ (slope $m_1=\frac{1}{t}$)

Normal $QN$ at $N=>y=\frac{x}{t}-\frac{a}{t^3}-\frac{2a}{t}$ (slope $m_2=\frac{1}{t}$)

$PT$ parallel $QN=>\alpha =0$

So, distance between $PT$ and $QN=\left|\frac{a{t}^2+2a+\frac{9}{t^2}}{\sqrt{1+t^2}}\right|$

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

$=>d=\left|\frac{a(1+t^2{)}^{\frac{3}{2}}}{t^2}\right|.$