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Ask community Community Discussion Question: consider a curve ax^2+2hxy+by^2=1 and a point P not on d curve. a line drawn from dis point interse
Reply Forum Index -> Analytical Geometry originally posted here on IIT-JEE / AIEEE community   
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TEJASWI V SHETTY. (169)

Olaaa!! Perrrfect answer. 29  [41 rates]

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consider a curve ax^2+2hxy+by^2=1 and a point P not on d curve. a line drawn from dis point intersects d curve at Q n R. If d product of PQ.PR is independent of d slope of d line,den show dat d curve is a circle.
    
~~~iShAli~~~ (54)

Olaaa!! Perrrfect answer. 8  [15 rates]

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let P be (m,n). parametric eq of line through P is

x=rcos(A)+m, y=rsin(A)+n

solving with the given curve

a(rcos(A)+m)+ b(sin(A)+n)2 +2h(rcos(A)+m)(rsin(A)+N) -1=0

frm this we get

PQ.PR=mod(r1)mod(r2)=mod(am2+bn2+2hmn-1)/(acos2(A) + asin2(A) - hsin(2A))..........(1)

if PQ.PR is constant then as the numerator in eq (1) is const. the denominator will also be const.

i.e. acos2(a)+bsin2(A) - hsin(2A) must be a constant for all values of (A) ....let that const be C.....

put (A)=o...we get a=C

(A)=90o....we get b=C

frm the abv eqtns we get  .......a=b.......(2)

substituting (2) in the denominator of eq 1we get

a-hsin(2A)=C

but a=C

so it gives hsin(2A)=0......or h=0............(3)

hence it is a circle from eq (2) and (3)

 

 

 

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