If the x-intercept of a focal chord of the pa | vedclass

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

Question

(B) The given equation of the parabola is $y^2 = 8x + 4y + 4$.Rewriting it by completing the square: $y^2 - 4y = 8x + 4$.$(y - 2)^2 - 4 = 8x + 4$,whic

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(B) The given equation of the parabola is $y^2 = 8x + 4y + 4$.Rewriting it by completing the square: $y^2 - 4y = 8x + 4$.$(y - 2)^2 - 4 = 8x + 4$,which simplifies to $(y - 2)^2 = 8(x + 1)$.Comparing this with the standard form $Y^2 = 4aX$,where $Y = y - 2$,$X = x + 1$,and $4a = 8$,we get $a = 2$.The vertex is $(-1, 2)$ and the focus is $(a - 1, 2) = (2 - 1, 2) = (1, 2)$.$A$ focal chord passes through the focus $(1, 2)$. Let the slope of the chord be $m$. The equation of the chord is $y - 2 = m(x - 1)$.Since the $x$-intercept of the chord is $3$,the point $(3, 0)$ lies on the line.Substituting $(3, 0)$ into the equation: $0 - 2 = m(3 - 1) \implies -2 = 2m \implies m = -1$.The length of a focal chord with slope $m$ for a parabola $Y^2 = 4aX$ is given by $L = 4a(m + \frac{1}{m})^2$ is incorrect; the correct formula is $L = 4a(1 + \frac{1}{m^2}) 	imes 	ext{something}$,but specifically for $m = -1$,the length is $4a(1 + \frac{1}{m^2}) = 4(2)(1 + \frac{1}{(-1)^2}) = 8(1 + 1) = 16$.

If the $x$-intercept of a focal chord of the parabola $y^2 = 8x + 4y + 4$ is $3$,then the length of this chord is equal to $.............$

Similar Questions

The length of the latus rectum of the parabola whose focus is $(3,3)$ and directrix is $3x - 4y - 2 = 0$ is . . . . . . units.

If $2x + y + \lambda = 0$ is a focal chord of the parabola $y^2 = -8x$,find the value of $\lambda$.

If the area of the triangle whose one vertex is at the vertex of the parabola,${y^2} + 4(x - {a^2}) = 0$ and the other two vertices are the points of intersection of the parabola and $y$-axis,is $250 \text{ sq. units}$,then a value of $a$ is

Statement $-1:$ The line $x - 2y = 2$ meets the parabola $y^2 + 2x = 0$ only at the point $(-2, -2).$
Statement $-2:$ The line $y = mx - \frac{1}{2m}$ $(m \neq 0)$ is tangent to the parabola $y^2 = -2x$ at the point $\left( -\frac{1}{2m^2}, -\frac{1}{m} \right).$

If the directrix of the parabola $x^2+4y-6x+\lambda=0$ is $y+1=0$,then which of the following is correct?

Vedclass Test Series

Exam Paper Generator

Online Exam Module