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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)

```
6 years ago

```
Hi,
The equation of the normal to the parabola y2 = 4ax is y = 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) then
cos 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 zero
So from (iii), 2a – sink <=0
=> 2 a <= sink
=> a <= sink/2
Since sink values lies between -1 and 1, and a >0, so a values will be between (0, ½).

So answer is choice  (2) .

```
6 years ago
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