Question

Find the locus of the midpoint of the portion of the straight line distance p from the origin intersect between the axes

Answer 1

SOLUTION

TO DETERMINE

The locus of the midpoint of the portion of the straight line distance p from the origin intersect between the axes

EVALUATION

Let the given equation of the line is

$\displaystyle \sf{ \frac{x}{a} + \frac{y}{b} = 1 } \: \: \: - - - - (1)$

The line given by Equation 1 cuts X axis at A(a, 0) & y axis at B(0,b) respectively

Let (h, k) be the midpoint of the line AB

Then

$\displaystyle \sf{h = \frac{a + 0}{2} \: \: \: and \: \: k = \frac{0 + b}{2} }$

$\displaystyle \sf{ \implies \: h = \frac{a }{2} \: \: \: and \: \: k = \frac{b}{2} }$

$\displaystyle \sf{ \implies \: a = 2h \: \: \: and \: \: b = 2k }$

Now the Perpendicular Distance from origin to the line AB is

$\displaystyle \sf{ \:p = \bigg| \frac{1}{\displaystyle \sf{ \sqrt{ \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } } }} \bigg| }$

$\displaystyle \sf{ \implies \frac{1}{ {a}^{2} } + \frac{1}{ {b}^{2} } = \frac{1}{ {p}^{2} } }$

$\displaystyle \sf{ \implies \frac{1}{ {(2h)}^{2} } + \frac{1}{ {(2k)}^{2} } = \frac{1}{ {p}^{2} } }$

$\displaystyle \sf{ \implies \frac{1}{ {h}^{2} } + \frac{1}{ {k}^{2} } = \frac{4}{ {p}^{2} } }$

Hence the required locus is

$\displaystyle \sf{ \frac{1}{ {x}^{2} } + \frac{1}{ {y}^{2} } = \frac{4}{ {p}^{2} } }$

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Answer 2

Answer:

p²(x²+y²)=4x²y²

Step-by-step explanation:

hope this answer helps you but i don't know explaination sorry