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The distance of vertex from the directrix is $\sqrt 2$

Further, the slope of the axis is $-1$

The focus is $(h,k)$

Then $$\frac{h-3}{\frac{-1}{\sqrt 2}}=\sqrt 2$$ $$h=2$$

And $$\frac{k-2}{\frac{1}{\sqrt 2}}=\sqrt 2$$ $$k=3$$

The slope of latus rectum is 1

$$y-3=x-2$$$$x-y=-1$$

The answer given is $x-y=3$

What’s going wrong?

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$h$ and $k$ are not computed correctly. They should be,

$$h= 3+ \frac1{\sqrt2}\sqrt2=4,\>\>\>\>k=2-\frac1{\sqrt2}\sqrt2=1$$

which yields $x-y=3$.

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  • $\begingroup$ Okay, I get that my computation was wrong, but how exactly? The angle was $\frac{3\pi}{4}$, so the cos component will be negative $\endgroup$
    – Aditya
    Commented Feb 25, 2020 at 15:44
  • $\begingroup$ @Aditya - the angle $\frac{3\pi}4$ is the direction from$(h,k)$ to $(3,2)$, which means $3-h=\sqrt2\cos\frac{3\pi}4$ and $h=4$. $\endgroup$
    – Quanto
    Commented Feb 25, 2020 at 16:40

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