Question

If $(h,k)$ is a point on the axis of the parabola $2(x-1{)}^2+2(y-1{)}^2=(x+y+2{)}^2$ from where three distinct normals may be drawn, then

• A. $h>2$
• B. $h>4$
• C. $h>8$
• D. $h<8$

Answer 1

The correct option is A $h>2$

We have,
$2(x-1{)}^2+2(y-1{)}^2=(x+y+2{)}^2$
$\Rightarrow \sqrt{(x-1{)}^2+(y-1{)}^2}=\left|\frac{x+y+2}{\sqrt{1+1}}\right|$
Clearly, it represents a parabola having its focus at $(1,1)$ and directrix $x+y+2=0$.
The equation of the axis is $y-1=1(x-1)$ ie, $y=x$.
Semi latusrectum $=$ Length of perpendicular from $(1,1)$ on the directix.
$\Rightarrow$ Semi latusrectum $=\left|\frac{1+1+2}{\sqrt{1+1}}\right|=2\sqrt{2}$
The coordinates of the vertex are $(0,0)$.
So, the equation of the axis in parametric form is
$\frac{x-0}{\cos \frac{\pi }{4}}=\frac{y-0}{\sin \frac{\pi }{4}}...\left(i\right)$
We know that three distinct normals can be drawn from a point $(h,0)$ on the axis of the parabola $y^2=4ax$,
if $h>2a(=\text{semi latusrectum})$.
The coordinates of a point on the axis $\left(i\right)$ at a distance $2\sqrt{2}$ from the vertex are given by
$\frac{x}{\cos \frac{\pi }{4}}=\frac{y}{\sin \frac{\pi }{4}}=2\sqrt{2}\Rightarrow x=2,y=2$