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Locus Problems
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The very important locus problems are discussing this video with examples.

## Sachin Rana is teaching live on Unacademy Plus

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Thnx for ur effort Debalina..it will surely help in revision. looking forward to it.
2 nd ques mein y(y-c) =4 kese ho gya?
Regarding the previous the HW solution we could also find the equation of hyperbola using y = mx + a/m
Aman Kumar
a month ago
yes, sir do not tell us about T^2=SS1
Aman Kumar
a month ago
I solve that question by using y =mx+ a/m
Locus of LP is a line parallel to the conjugate axis i.e. the y-axis. Equation is x= asec(theta) which is a line, which itself passes through the assumed point P on the hyperbola.
Mohit Likhar
a year ago
2nd question Mai locus kese aaya please explain to me
As assumed initially.. x1,y1 and x2,y2 are the co-ordinates that lies on the hyperbola.. so we are bringing the value of x1 and y1 in terms of h and k.. now we have x1 and y1 satisfying the equation xy=1 as it lies on it.. therefore we can substitute the values obtained in this equation to obtain the locus by solving further. (i.e. x1=-h-k/2 , y1=-k-8h ).
Mohit Likhar
a year ago
thank you so much
Anytime :-)
Line is parallel to conjugate axis with equation x=asec¥ ( if we consider any point pas (asec¥, btan¥). and point G(0,t) then point L is [(-tab/b^2+a^2) , -t(a^2/b^2+a^2 - 1)] which gives slope of line LP=0
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