If you can manage to explain why, I will be amazed (1 Viewer)

[wagig]

For Question 6.b) ii) of the 2008 HSC exam, you would find the perpendicular distance from the focus to the line, and as a result you get an absolute value in the numerator. Why would you remove the absolute value?

 

[RealiseNothing]

Both and

 

[wagig]

You Sir are a genius! You have opened my eyes to the world around me!
Thanking you!

 

[wagig]

Here have a doge coin *gives doge coin*

 

[QZP]

Both and

why is sec theta > 1? I thought diagrams were just like... guidelines. Just because P is in first quadrant shouldn't mean there is a restriction on theta?

 

[Sy123]

why is sec theta > 1? I thought diagrams were just like... guidelines. Just because P is in first quadrant shouldn't mean there is a restriction on theta?

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[iStudent]

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I think QZP is saying what if theta is between 90 and 270
Because then sec(theta) becomes negative

 

[Sy123]

I think QZP is saying what if theta is between 90 and 270
Because then sec(theta) becomes negative

Oh, well in that case the diagram shows that P varies on one arm of the hyperbola, meaning that -90 < theta < 90 meaning sec is positive.

 

[braintic]

Oh, well in that case the diagram shows that P varies on one arm of the hyperbola, meaning that -90 < theta < 90 meaning sec is positive.

Would you please stop that damn frog from moving. I'm wondering what is going on below the viewing window.