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 can be drawn, then prove that `h > 2.`

Options

A. `hgt2`
B. `hlt4`
C. `hgt8`
D. `hlt8`

Answer

A

Explanation

We have,
`2(x-1)^(2)+(y-1)^(2)=(x+y+2)^(2)`
`rArr" "sqrt((x-1)^(2)+(y-1)^(2))=|(x+y+2)/(sqrt(1+1))|^(2)`
Clearly, it represents a parabola having its focus at (1, 1) and directrix x + 2 = 0.
The equation of the axis is `y-1(x-1) i.e. y = x`
Semi-latusrectum = Length of `bot` from (1, 1) on the directrix
`rArr" Semi-laturectum "=|(1+1+2)/(sqrt(1+1))|=2sqrt2`.
The coordinates of the vertex are (0, 0).
So, the equation of the axis in parametrix form if
`(x-0)/("cos "pi//4)=(y-0)/("sin "pi//4)" ...(i)"`
We know that three distinct normals can be drawn from a point (h, 0) on the axis of the parabola `y^(2)=4ax` if
`hgt2a("= semi-latusretum")`.
The coordinates of a point on the axis (i) at a distance `2 sqrt2` from the vertex are given by
`x/("cos "pi//4)=y/("sin "pi//4)=2sqrt2rArrx=2, y=2`.
Hence, three normals can be drawn from (h, k), if `hgt2`.