Question

If a line passing through a point P(1,0) and having slope '1' meets the parabola $y^2=x$ in 'A' and 'B' then |PA| + |PB| is

• A. $3\sqrt{2}$
• B. $\sqrt{8}$
• C. $\sqrt{10}$
• D. None of these

Answer 1

Answer: (C)

$\sqrt{10}$

If line passing through a point $P(1,0)$ and having slope $1$ meets the parabola $y^2=x$ in $A$ and $B$.

The to find $\left|PA\right|+\left|PB\right|$

Equation of given line is

$y-0=1(x-1)y=x-\mathrm{1.........}\left(1\right)y^2=x...........\left(2\right)$

Solving (1) and (2)

$x^2+1-2x=x{x}^2-3x+1=0x=\frac{3\pm \sqrt{9-4}}{2}=\frac{3\pm \sqrt{5}}{2}y=\frac{1\pm \sqrt{5}}{2}A(\frac{3+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2})B(\frac{3-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})PA=\sqrt{{(\frac{3+\sqrt{5}}{2}-1)}^2+{(\frac{1+\sqrt{5}}{2}-0)}^2}=\frac{1+\sqrt{5}}{\sqrt{2}}PB=\sqrt{{(\frac{3-\sqrt{5}}{2}-1)}^2+{(\frac{1-\sqrt{5}}{2}-0)}^2}=\frac{1-\sqrt{5}}{\sqrt{2}}\left|PA\right|+\left|PB\right|=\frac{1+\sqrt{5}}{\sqrt{2}}+\frac{1-\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}$