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The equation of a parabola with the line $\mathrm{x+y=2 }$ as tangent at the vertex and focus at the point $\mathrm{(4,4)}$ is

$\mathrm{\mathrm{k}\left[(\mathrm{x}-4)^2+(\mathrm{y}-4)^2\right]=(\mathrm{x}+\mathrm{y}+4)^2}$ , where $\mathrm{\mathrm{k}=}$
Option: 1
$1$

Option: 2
$2$

Option: 3
$3$

Option: 4
$\frac{1}{2}$

Answers (1)

best_answer

Let $\mathrm{S(4,4)}$ be the focus and V be the vertex of the parabola.

Equation of the axis of the parabola is

$\mathrm{ \begin{aligned} & y-4=(+1)(x-4) \Rightarrow y-4=x-4 \\ & \Rightarrow y=x \end{aligned} }.......(1)$

Solving (1) with $\mathrm{ x+y=2 }$

We get $\mathrm{ V=(1,1)}$

Let co-ordinates of D be $\mathrm{ (\alpha, \beta) }$ , then $\mathrm{ \frac{\alpha+4}{2}=1}$ and

$\mathrm{ \frac{\beta+4}{2}=1}$

$\mathrm{ \Rightarrow \alpha=-2 \text { and } \beta=-2}$

Thus co-ordinates of D are $\mathrm{(-2,-2)}$

Now equation of directrix can be written as $\mathrm{ x+y=\lambda}$

Since it passes through $\mathrm{ (-2,-2)}$ hence,

$\mathrm{ \lambda=-2-2=-4 }$

Thus equation of directrix is $\mathrm{x+y=-4}$

So equation of parabola is

$\mathrm{ \sqrt{(x-4)^2+(y-4)^2}=\frac{|x+y+4|}{\sqrt{2}} }$

or $\mathrm{2\left[(x-4)^2+(y-4)^2\right]=(x+y+4)^2}$

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