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Sachin Rana

IVth Year UG, IIT Bombay | YouTuber (143k subs) | Mentored 3 under 100 ranks in JEE Advanced | No. 1 educator for Organic Chemistry.

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Chandra Bhal Tiwari Om

3 years ago

Thnx for ur effort Debalina..it will surely help in revision. looking forward to it.

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Eshu Manohare

6 months ago

2 nd ques mein y(y-c) =4 kese ho gya?

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Codeck Unavailable

a year ago

Regarding the previous the HW solution we could also find the equation of hyperbola using y = mx + a/m

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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

Abantika Karmakar

a year ago

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.

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Sachin Rana

a year ago

Correct. Good.

Mohit Likhar

a year ago

2nd question Mai locus kese aaya please explain to me

Abantika Karmakar

a year ago

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

Abantika Karmakar

a year ago

Anytime :-)

Swaraj

a year ago

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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