The ends of the latus rectum of the conic {x^ | vedclass

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

Question

(C) The given equation of the conic is ${x^2} + 10x - 16y + 25 = 0$.Rewriting the equation by completing the square: ${(x + 5)^2} - 25 - 16y + 25 = 0$

Vedclass · Accepted Answer

$(3, 4), (-13, 4)$

Vedclass · Answer

(C) The given equation of the conic is ${x^2} + 10x - 16y + 25 = 0$.Rewriting the equation by completing the square: ${(x + 5)^2} - 25 - 16y + 25 = 0$,which simplifies to ${(x + 5)^2} = 16y$.This is a parabola of the form ${(x - h)^2} = 4a(y - k)$,where $h = -5$,$k = 0$,and $4a = 16$,so $a = 4$.The focus of the parabola is $(h, k + a) = (-5, 0 + 4) = (-5, 4)$.The latus rectum is a horizontal line segment passing through the focus $(-5, 4)$ at $y = 4$.Substituting $y = 4$ into the equation ${(x + 5)^2} = 16(4) = 64$,we get $x + 5 = \pm 8$.Thus,$x = -5 + 8 = 3$ and $x = -5 - 8 = -13$.The endpoints of the latus rectum are $(3, 4)$ and $(-13, 4)$.

The ends of the latus rectum of the conic ${x^2} + 10x - 16y + 25 = 0$ are

Similar Questions

If the equation of a system of parallel chords of the parabola $y^2 = \frac{25x}{7}$ is $4x - y + \lambda = 0$,then the equation of the corresponding diameter is . . . . . .

The equation of the tangent to the parabola $y^2=16x$,which is perpendicular to the line $3x-4y+5=0$,is given by

The area of the triangle formed by the lines joining the vertex of the parabola $x^{2}=12y$ to the extremities of its latus rectum is

If $x+y=k$ is a normal to the parabola $y^{2}=12x$,then the value of $k$ is:

$STATEMENT-1$: The curve $y = -\frac{x^2}{2} + x + 1$ is symmetric with respect to the line $x = 1$. Because
$STATEMENT-2$: $A$ parabola is symmetric about its axis.

Vedclass Test Series

Exam Paper Generator

Online Exam Module