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If p is the length of the perpendicular from the origin to the line #x/a+y/b=1#,prove that #1/(p²)=1/(a²)+1/(b²)#?

Answer 1

Jayden Jackson

See below.

Considering the triangle

#BOA# with

#B = (0,b)# #O=(0,0)# #A = (a,0)#

we have

#bar(OB)^2-p^2=n^2# #bar(OA)^2-p^2 = m^2# #p^2=n m#

then

#(bar(OB)^2-p^2)(bar(OA)^2-p^2)=n^2m^2=p^4#

developping

#bar(OB)^2bar(OA)^2-bar(OB)^2 p^2-bar(OA)^2 p^2=0# or

#p^2 = (bar(OB)^2bar(OA)^2)/(bar(OA)^2+bar(OB)^2)# or finally

#1/p^2 = 1/bar(OA)^2+1/bar(OB)^2 = 1/a^2+1/b^2#

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Answer 2

Noah Atkinson

To prove: ( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} )

Given: Line equation ( \frac{x}{a} + \frac{y}{b} = 1 )

The distance from the origin to a line is given by the formula:

[ p = \frac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} ]

Here, ( x_1 = 0 ), ( y_1 = 0 ), and ( c = 0 ).

[ p = \frac{|0 + 0 + 0|}{\sqrt{a^2 + b^2}} = \frac{0}{\sqrt{a^2 + b^2}} = 0 ]

Therefore, ( p = 0 ).

Now, ( \frac{1}{p^2} = \frac{1}{0^2} = \text{undefined} ).

Similarly, ( \frac{1}{a^2} + \frac{1}{b^2} = \frac{b^2 + a^2}{a^2b^2} ).

Since ( p = 0 ) and the denominator cannot be zero, the equation ( \frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} ) is invalid. Therefore, the statement cannot be proved.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.