From a point (sin k,cos k),three normal can be dra

Question

From a point (sin k,cos k),three normal can be drawn to the parabola y^2=4ax(a>0) then then value of a lies in the interval:
1)(2,5/2)
2)(0,1/2)
3)(2,3)
4)(1/2,3/2)

VENNA RAMANJANEYA REDDY · Accepted Answer

Hi,The equation of thenormal to the parabolay2= 4ax isy = mx – 2am – am3, where m is the slope of the tangent line corresponding to the normal.If the normal passes through the point (sin k, cos k) thencos k = m sink – 2am – am3=> am3+ m(2a – sin k) + cosk = 0. … (i)Let the roots of the above equation be m1, m2and m3.From (i),m1+m2+m3= 0 --- (ii)m1 * m2 + m2 * m3+ m3 * m1 = 2a- sin k --(iii)m1 * m2 * m3= -cos k ----(iv)If sum of the roots = 0, then sum of the products of the roots will be less than or equal to zeroSo from (iii), 2a – sink <=0=> 2 a <= sink=> a <= sink/2Since sink values lies between -1 and 1, and a >0, so a values will be between (0, ½).So answer is choice (2) .

Analytical Geometry> From a point (sin k,cos k),three normal c...

From a point (sin k,cos k),three normal can be drawn to the parabola y^2=4ax(a>0) then then value of a lies in the interval:\n1)(2,5/2)\n2)(0,1/2)\n3)(2,3)\n4)(1/2,3/2)

Sameer Gupta , 10 Years ago

VENNA RAMANJANEYA REDDY

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Analytical Geometry> From a point (sin k,cos k),three normal c...

From a point (sin k,cos k),three normal can be drawn to the parabola y^2=4ax(a>0) then then value of a lies in the interval:\n1)(2,5/2)\n2)(0,1/2)\n3)(2,3)\n4)(1/2,3/2)

Sameer Gupta , 10 Years ago

VENNA RAMANJANEYA REDDY

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