Question

Consider the parabola whose focus is at $(0,0)$ and tangent at vertex is $x-y+1=0.$ Then

• A. the length of latus rectum is $2\sqrt{2}$
• B. the length of the chord of parabola on the $x$-axis is $4\sqrt{2}$
• C. equation of directrix is $x-y-2=0$
• D. tangents drawn to the parabola at the extremities of the chord $3x+2y=0$ intersect at an angle of $\frac{\pi }{3}$

Answer 1

Answer: (A), (B)

the length of the chord of parabola on the $x$-axis is $4\sqrt{2}$

The distance between the focus and the tangent at the 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 be $x-y+\lambda =0,$
where $\frac{\lambda }{\sqrt{1^2+1^2}}=\frac{2}{\sqrt{2}}$
$\Rightarrow \lambda =2$
$\Rightarrow$ Equation of directrix is $x-y+2=0$

Let $P(x,y)$ be any moving point on the parabola.
Then $OP=PM$
$\Rightarrow x^2+y^2={\left(\frac{x-y+2}{\sqrt{1^2+1^2}}\right)}^2$
$\Rightarrow 2x^2+2y^2=(x-y+2{)}^2$
$\Rightarrow x^2+y^2+2xy-4x+4y-4=0...\left(1\right)$


Latus rectum length $=2\times$ (distance of focus from directrix)
$=2\left|\frac{0-0+2}{\sqrt{1^2+1^2}}\right|$
$=2\sqrt{2}$

Now, solving parabola with $x$-axis by putting $y=0$ in equation $\left(1\right).$
$x^2-4x-4=0$
$\Rightarrow x=\frac{4\pm \sqrt{32}}{2}=2\pm 2\sqrt{2}$
$\Rightarrow$ Length of chord on $x$-axis is $4\sqrt{2}$.

Since, the chord $3x+2y=0$ passes through the focus, so it is focal chord.
Hence, tangents at the extremities of chord are perpendicular.