Question

Tangent are drawn at those point on the parabola $y^2 = 16x$ whose ordinate are in the ratio 4 : 1. If the locus of point of intersection of these tangents is $y^2 = kx$, then $[k/3]$ is: ([.] denote greatest integer function,

Answer 1

8

Answer 2

The given parabola is $y^2 = 16x$. This is of the form $y^2 = 4ax$, where $4a = 16$, so $a = 4$.

Let the two points on the parabola be $P_1$ and $P_2$. In parametric form, a point on the parabola $y^2=4ax$ is $(at^2, 2at)$. For $y^2=16x$, the parametric points are $(4t^2, 8t)$. Let the two points be $P_1(4t_1^2, 8t_1)$ and $P_2(4t_2^2, 8t_2)$.

The ordinates of these points are $y_1 = 8t_1$ and $y_2 = 8t_2$. Given that their ordinates are in the ratio 4:1:

$\frac{y_1}{y_2} = \frac{4}{1}$

$\frac{8t_1}{8t_2} = \frac{4}{1}$

$\frac{t_1}{t_2} = \frac{4}{1} \implies t_1 = 4t_2$.

The coordinates of the point of intersection of tangents drawn at $(at_1^2, 2at_1)$ and $(at_2^2, 2at_2)$ are $(at_1t_2, a(t_1+t_2))$. For our parabola, $a=4$. Let the point of intersection be $(h, k)$.

So, $h = 4t_1t_2$

And $k = 4(t_1+t_2)$

Substitute $t_1 = 4t_2$ into these expressions:

$h = 4(4t_2)t_2 = 16t_2^2$ (1)

$k = 4(4t_2+t_2) = 4(5t_2) = 20t_2$ (2)

To find the locus of $(h, k)$, we need to eliminate the parameter $t_2$. From equation (2), $t_2 = \frac{k}{20}$. Substitute this value of $t_2$ into equation (1):

$h = 16 \left(\frac{k}{20}\right)^2$

$h = 16 \frac{k^2}{400}$

$h = \frac{k^2}{25}$

Rearranging the equation to match the standard form of a parabola:

$k^2 = 25h$

Replacing $(h, k)$ with $(x, y)$ to represent the locus:

$y^2 = 25x$

The problem states that the locus of the point of intersection of these tangents is $y^2 = kx$. Comparing our derived locus $y^2 = 25x$ with $y^2 = kx$, we find:

$k = 25$

Finally, we need to find the value of $[k/3]$, where $[.]$ denotes the greatest integer function.

$[k/3] = [25/3]$

$[25/3] = [8.333...]$

The greatest integer less than or equal to 8.333... is 8.

So, $[k/3] = 8$.