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Grade: 11

                        

The value of k for which the line x+y+1=0 touches the parabola y^2=kx is

3 years ago

Answers : (3)

Saurabh Koranglekar
askIITians Faculty
8407 Points
6 months ago
Vikas TU
12281 Points
							
 
The line x+y=1 touches the parabola y^2=kx or y^2=4.k/4 x ( Compare with standard eqn y^2=4ax) so a= k/4 and the condition of tangency for the line y=mx+c  to a parabola y^2=4ax is that : c=a/m
we ghave given c=1,  m=-1, a=k/4 This implies 1=k/4(-1) => k=-4
6 months ago
Arun
24735 Points
							

x+y-1=0. or. y=1-x………(1)

y^2. = kx……………………..(2)

On putting y=1-x from eq.(1)

(1-x)^2=k.x

1–2x+x^2 =kx

x^2 - (k+2)x+1 = 0

The given line is tangent to the parabola which touches the parabola one and only one point. means both the roots are same.

D or. b^ - 4 a.c. =0

{-(k+2)}^2. - 4.1.1. =0

(k+2)^2. = 4

(k+2) = +/-(2)

k= -2 +/-2

k= -2+2 or. -2–2.

k =0. , - 4. Answer.

6 months ago
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