Question

The focus of the parabola is $(1,1)$ and the tangent at the vertex has the equation $x+y=1$. Then:

• A. equation of the parabola is ${(x-y)}^2=2(x+y-1)$
• B. equation of the parabola is ${(x-y)}^2=4(x+y-1)$
• C. the coordinates of the vertex are $(\frac{1}{2},\frac{1}{2})$
• D. length of the latus rectum is $2\sqrt{2}$

Answer 1

The correct option is B equation of the parabola is ${(x-y)}^2=4(x+y-1)$

Let $S$ be the focus and $A$ be the vertex of the parabola. let K be the point of intersection of the axis and directrix.

since axis is a line passing through $S(1,1)$ and perpendicular to $x+y=1.$

So, let the equation of the axis be,

$x-y+\lambda =0$

This will pass through $(1,1)$ if,

$1-1+\lambda =0\Rightarrow \lambda =0$

So, the equation of the axis is $x-y=0$

The vertex $A$ is the point of intersection of $x-y=0$ and $x+y=1.$

Solving these two equations, we get

$x=\frac{1}{2}$ and $y=\frac{1}{2}$

So, the coordinates of the vertex $A$ are $(\frac{1}{2},\frac{1}{2})$.

let $(x_1,y_1)$ be the co-ordinates of $k.$ Then,

$\Rightarrow$ $\frac{x_1+1}{2}=\frac{1}{2},\frac{y_1+1}{2}=\frac{1}{2}$

$\Rightarrow$ $x_1=0,$ $y_1=0$

So, the co-ordinates of $K$ are $(0,0).$

Since directrix is a line passing through $K(0,0)$ and parallel to $x+y=1.$

$\therefore$ Equation of the directrix is

$y-0=-1(x-0),$ $i.e.x+y=0$

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

Then, distance of $P$ from the focus $S=$ distance of $P$ from the directrix.

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

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

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

$\Rightarrow$ $(x-y{)}^2=4x+4y-4$

$\Rightarrow$ $(x-y{)}^2=4(x+y-1)$