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Chandrashekhar Wankhede 7 years, 10 months ago

Any point on Parabola will have co-ordinates (at2, 2at) read as (a t square, 2 a t) . where t is unique for a point on parabola. It is given that all three vertices of equilateral triangle are lying on parabola y2= 4ax. One of point is O(0,0). Other two points will be at t = t1 --> P(at12, 2at1) , at t = t2 --> Q(at22, 2at2). As all three points are vertices of equilateral triangle then OP = OQ. Solving by distance formula, we will get t12= t22 ( read as t1 square = t2 square) , which mean t1= t2, t1= -t2 or -t1= t2. t1 can't be equal to t2 otherwise triangle can't be formed. It means t1 = - t2 , or - t1= t2 , it mean Y co-ordinates(2at) of other two points will have same magnitude but opposite signs. It means that axis of parabola bisects the side of equilateral triangle opposite to vertex (0,0). Also Distance PQ will be length of side of equilateral triangle.
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Chandrashekhar Wankhede 7 years, 10 months ago

Pl clear " vowels are consonents" ..Retype correct question..
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Yashu Tyagi 7 years, 10 months ago

4x=2×2x let 2x=∆ now, cos2∆=1-2sin"∆ put value of ∆=2x cos4x =1-2(sin 2x)". sin2x =2sinxcosx cos4x =1-2(2sinxcosx)" cos4x =1-2(4sin"xcos"x) cos4x =1-8sin"xcos"x
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Chandrashekhar Wankhede 7 years, 10 months ago

Normal form of Line 3x-4y=-4 is , x cos k + y Sin k = p, where p is length of perpendicular from origin to that line and k is angle made by that perpendicular with positive x-axis. As both forms represents same line, Alternately we can consider them as two different coinciding lines. Hence we can write 3/(cos k) = -4/(sin k) = -4/p From this equation we can determine cos k and sin k and then value of p by using identity sqr(Sin k) + sqr(cos k)= 1. You can also determine tan k. As per my calculation tan k = -4/3 therefore angle k = tan inverse (-4/3) , angle k will be greater than 90 degree as tan k is negative. Also p = 4/5.. We can directly substitute cos k and sin k values in normal form x cos k + y Sin k = p, without knowing actual value of k.

Chandrashekhar Wankhede 7 years, 10 months ago

x cos $ + y sin $ = p is normal form of same line 3x - 4y = - 4 , where p is perpendicular distance of line from origin and $ is angle made by that perpendicular with positive X axis. As both equation represents same line we can consider them as two different lines which are "coinciding" . Hence we can write (3/ cos $)=(-4/sin $)=(-4/p) , from these ratios we can find cos $, sin $ and hence tan $ and angle $. p can also determined using identity sqr(sin $) + sqr(cos $) = 1. After knowing value of p and $ , we can substitute in normal form equation..
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Rishabh Dixit 7 years, 10 months ago

1

Ashutosh Kumar Jha 7 years, 11 months ago

1

Murli Pandey 7 years, 11 months ago

1 since it's value is formed by division of sin45 and cos45 which is 1/2*2=1

Ankita Karn 7 years, 11 months ago

1
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Chirag Bhardwaj 7 years, 11 months ago

In mathematical operations.
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Mayank Pathak 7 years, 11 months ago

Dy/dx=x×d(sinx)/dx+sinxd(x)/dx = Xcosx+sinx..... Because d(sinx)/dx=cosx anddx/dx=1
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Rajnish Kumar 7 years, 11 months ago

Is the ans 1. ???
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Anjali Jaiswal 7 years, 10 months ago

Lim sin^2x+h-sin^2x upon h htending to0 Lim. Sin2x+h.sinh upon h. (sin^2A-sin htending to 0. =sin2x

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