Question

Consider the parabola whose focus is at (0,0) and tangent at vertex is $x-y+1=0$
The length of latus rectum is

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $8\sqrt{2}$
• D. $3\sqrt{2}$

Answer 1

Answer: (B)

$2\sqrt{2}$

The distance between the focus and the tangent at the

$\text{vertex}=\frac{|0-0+1|}{\sqrt{1^2+1^2}}=\frac{1}{\sqrt{2}}$

The directrix is the line parallel to the tangent at vertex and at a distance $2\times \frac{1}{\sqrt{2}}$ from the focus.

Let equation of directrix is

$x-y+\lambda =0$

$\text{where}\frac{\lambda }{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}$

$\Rightarrow \lambda =2$

Let $P(x,y)$ be any moving point on the parabola, then

$OP=PM$

$\Rightarrow x^2+y^2={\left(\frac{x-y+2}{\sqrt{1^2+1^2}}\right)}^2$

$\Rightarrow 2x^2+2y^2=(x-y+2{)}^2$

$\Rightarrow x^2+y^2+2xy-4x+4y-4=0$

Latus rectum length $=2\times$ (distance of focus from directrix )

$=2\left|\frac{0-0+2}{\sqrt{1^2+1^2}}\right|$

$=2\sqrt{2}$

Solving parabola with $x$ -axis,

$x^2-4x-4=0$

$\Rightarrow x=\frac{4\pm \sqrt{32}}{2}=2\pm 2\sqrt{2}$

$\Rightarrow$ Length of chord on $x$ -axis is $4\sqrt{2}$

Since the chord $3x+2y=0$ passes through the focus, it is focal chord.

Hence, tangents at the extremities of chord are perpendicular.