The straight line y = 2x + lambda does not me | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The given line is $y = 2x + \lambda$ and the parabola is $y^2 = 2x$.Substituting $y = 2x + \lambda$ into the parabola equation: $(2x + \lambda)^2

Vedclass · Accepted Answer

$\lambda > \frac{1}{4}$

Vedclass · Answer

(B) The given line is $y = 2x + \lambda$ and the parabola is $y^2 = 2x$.Substituting $y = 2x + \lambda$ into the parabola equation: $(2x + \lambda)^2 = 2x$.$4x^2 + 4x\lambda + \lambda^2 = 2x$.$4x^2 + (4\lambda - 2)x + \lambda^2 = 0$.For the line not to meet the parabola,the discriminant $D$ must be less than $0$.$D = (4\lambda - 2)^2 - 4(4)(\lambda^2) $16\lambda^2 - 16\lambda + 4 - 16\lambda^2 $-16\lambda + 4 $16\lambda > 4$.$\lambda > \frac{4}{16} = \frac{1}{4}$.

The straight line $y = 2x + \lambda$ does not meet the parabola $y^2 = 2x$,if

Similar Questions

Find the equation of the tangent to the curve $y^2 = 6x$ at the point $(2, -3)$.

If the line $x + y - 1 = 0$ is tangent to the parabola $y^2 = kx$,find the value of $k$.

If $a$ and $c$ are the segments of a focal chord of a parabola and $b$ is the semi-latus rectum,then:

For the parabola $y=2+4t, x=-2+2t^2$,the ends of the latus rectum are at $t=\alpha$ and $t=\beta$. Then $\alpha \beta=$

If $S(a, b)$ is a fixed point and $P(\alpha, \beta)$ is a variable point such that $4[(x-a)^2+(y-b)^2]=(\alpha x+\beta y+7)^2$ represents a parabola,then the locus of $P(\alpha, \beta)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module