Question

If the line x+y-1=0 touches the parabola y^2=kx then the value of k is

Answer 1

they will satisfy each other
x+y=1
y=1-x ...........................(1)
and y^2 =kx
from (1)
(1-x) ^2=kx
(1+x^2 -2x)/x =k

Answer 2

Answer:

value of k = -4

Step-by-step explanation:

As per the statement:

If the line x+y-1=0 touches the parabola $y^2=kx$

Find the value of k:

We can write the equation of line as:

$y = 1-x$

Substitute the value of y in parabola equation we have;

$(1-x)^2= kx$

Using the identity rule:

$(a-b)^2 = a^2-2ab+b^2$

then;

$1+x^2-2x=kx$

then;

$x^2-2x-kx+1 = 0$

⇒$x^2+(-2-k)x+1=0$

Since, the line touches the parabola if both real roots of the above quadratic equation are equal,

we have the determinant(D) is given by:

$D = b^2-4ac$ = 0

On comparing given equation with $ax^2+bx+c = 0$ we have;

a = 1, b = -2-k and c = 1

then;

$(-2-k)^2-4(1)(1)=0$

⇒$(-2-k)^2 = 4$

⇒$-2-k = \pm 2$

then;

$-2-k = -2$  or $-2-k = 2$

⇒$k = 0$ or k = -4

For k = 0, y^2=kx is not a parabola.

neglect k = 0

Therefore, the value of k is, -4