Question

Let $AB$ be a focal chord (not the latus rectum) of the parabola $y^2=4px$ where $p$ is a prime number$(p>0),$ such that the lengths, $SA$ and $SB$ are integers($S$ being the focus of the parabola). If the tangents to the parabola at $A$ and $B$ meet at point $P,$ then

• A. $PA=(p+1)\sqrt{p+1}$
• B. $PB=(p+1)\sqrt{p(p+1)}$
• C. $SP=(p+1)\sqrt{p}$
• D. $AB=(p+1{)}^2$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A

$PA=(p+1)\sqrt{p+1}$

B

$PB=(p+1)\sqrt{p(p+1)}$

C

$SP=(p+1)\sqrt{p}$

D

$AB=(p+1{)}^2$

Let $SA=a,SB=b,$ then $\frac{1}{a}+\frac{1}{b}=\frac{1}{p}$

$=\frac{a+b}{ab}=\frac{1}{p}$

$=p(a+b)=ab$

$\Rightarrow (a-p)(b-p)=p^2\Rightarrow a=p+1,b=p+p^2$

So, $AB=a+b=p^2+2p+1=(p+1{)}^2$

Since, $AB$ is the focal chord, $P$ lies on the directrix and the tangents, so, $PA$ and $PB$ are perpendicular.

$\Rightarrow SP=\sqrt{SA\times SB}=\sqrt{(p+1)(p+p^2)}=(p+1)\sqrt{p}$

$PA=\sqrt{S{P}^2+S{A}^2},$ and $PB=\sqrt{S{P}^2+S{B}^2}$

$\Rightarrow PA=\sqrt{(p+1{)}^{2}p+(p+1{)}^2}=(p+1)\sqrt{p+1}$

and $PB=\sqrt{(p+1{)}^{2}p+p^2(p+1{)}^2}=(p+1)\sqrt{p(p+1)}$