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Call the intersection of PF and DQ as G. Since Ang GFE+ Ang GDE = 180 0 , GFED is a cyclic quad and hence G also lies on the incircle. Now join FD. Let Ang FED = b. Then it easily follows that Ang...
Eqn. of linee joining these two points is: (y – 6) = ((-3-6)/(2-5))(x – 5) y – 6 = (-9/-3)(x – 5) y – 6 = 3x – 15 y = 3x – 9 when y = 0 3x = 9 x = 3 at x = 3 it cuts the x – axis.
but the above answer is true if and only if OAPB is a cyclic quadrilateral ?? how do we know it is cyclic or not ??
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Since the circumcircle of ∆ PAB, will have the diameter joining P and the centre of the circle x^2+y^2=a^2.i.e.( 0,0)Therefore , the eq. Of the circumcircle is [ x( x-x1) + y(y-y1). ]
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Let that point be (x,y) Then from distance formulae: 4 = (x-5) ^2 + (y-4)^2.............(1) 100 = (x-11)^2 + (y+2)^2...............(2) Solving the eqns. (1) and (2) we get, x – y + 1 = 0 or the...
Hii The most important suggestion for complete geometry is to draw the figure of the questiob before starting it. In this way you will be able to find out the progress of yours in the test.
@nikhil solution book are avaliable separately for a das gupta books in the market , its of bharat publication . also , if u have questions , then u can directly post it here , and i will answer that ...
tan45 = |3 – m|/|1 + 3m| = 1/root2 solvinf for m we get, m = ½. TANGENT OF parabola = > 2ydy/dx = 8 dy/dx = 4/y1 |let x1 and y1 be the point of contact] m = 4/y1 = ½ y1 = 8. x1 = 64/8 = 8 the eqn....
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