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Analytical Geometry

Cool goIITian

Joined:	19 Aug 2007
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24 Oct 2007 22:33:28 IST

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hurrrrrrraaayyyyyyyyyyyy answer

None

A line cuts the X axis at A(7,0) & the Y axis at B(0,-5).A variable PQ is drawn perpendicular to AB cutting the X axis at P and the Y axis at Q .If AQ & BP intersects at R,find the locus of R.

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Rohan

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24 Oct 2007 22:48:06 IST

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x^2 + y^2 - 7x + 5y = 0

am i right ??

Rohan

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24 Oct 2007 23:08:21 IST

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tell me if it is right and i will post the solution

sankydreams

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24 Oct 2007 23:22:26 IST

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Let P(p,0), Q(0,q),R(h,k)
eqn of AB is x/7 - y/5 = 1
slope = 5/7
=>slope of PQ is -7/5
q/p = -7/5
q = -7p/5..............(1)

eqn of AQ
x/7 + y/q = 1
eqn of BP
x/p - y/5 = 1

solving these for y
y = (35 - 5p) / (p+5) = k
solve for x also and put x = h
Then eliminate p and q using (1)
Rate if this helps u.........

Rohan

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24 Oct 2007 23:30:33 IST

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my solution goes like this:::

any line through (7,0) and (0,-5) is:::

x/7 + y/-5 = 1

PQ is prependicular to this line so it's equation is

x/5 + y/7 = a (let it be)
or
x/5a + y/7a = 1

now the equation of AQ is

x/7 + y/7a = 1
or
y=a(7-x).....................................1

eqn of BP

x/5a - y/5 = 1
or
x=a(y+5).......................................2

dividing (1) and (2) to eliminate a (parameter)

we get
x^2 + y^2 - 7x + 5y = 0

which is the answer

don't u think i deserve a rate

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