Question

The parabola y^2 = kx makes an intercept of length 2√10 on the line x - 2y =1. Then k is :
(A) 1
(B) -1
(C) 2
(D)-2
Step by Step Calculation Required.​

Answer 1

Answer:

See below.

Step-by-step explanation:

First, the given line will intersect the parabola in two points. We'll first find these two points.

We are given the line,

$\begin{aligned}L:x-2y-1=0\\ x=1+2y\end{aligned}$

Substitute this value of x in the equation of the parabola,

$\begin{aligned}P:y^{2}=kx\\ y^{2}=k\left( 1+2y\right) \\ y^{2}-2ky-k=0\end{aligned}$

The two roots of the above quadratic equation are,

$y=k\pm \sqrt {k\left( 1-k\right) }$

For these values of the ordinate, we get the two corresponding value of abscissas. The points at which the line intersects the parabola is,

$\begin{aligned}A:\left( 1+2k+2\sqrt {k\left( 1+k\right) },k+\sqrt {k\left( 1+k\right) }\right) \\ B:\left( 1+2k-2\sqrt {k\left( 1+k\right) },k-\sqrt {k\left( 1+k\right) }\right) \end{aligned}$

We are given the distance between these two points which is 2√10,

$\begin{aligned}\left( 4\sqrt {k\left( 1+h\right) }\right) ^{2}+\left( 2\sqrt {k( 1+k}\right) ^{2}=40\\ 20k\left( 1+k\right) =40\\ k^{2}+k-2=0\\ k=1,-2\end{aligned}$