If the directrix of the parabola x^2+4y-6x+la | vedclass

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

Question

(C) The given equation is $x^2-6x+4y+\lambda=0$. Completing the square for $x$: $(x-3)^2-9+4y+\lambda=0$,which simplifies to $(x-3)^2 = -4y + (9-\lamb

Vedclass · Accepted Answer

focus is $(3,-3)$

Vedclass · Answer

(C) The given equation is $x^2-6x+4y+\lambda=0$. Completing the square for $x$: $(x-3)^2-9+4y+\lambda=0$,which simplifies to $(x-3)^2 = -4y + (9-\lambda) = -4(y - \frac{9-\lambda}{4})$. Comparing this with the standard form $(x-h)^2 = -4a(y-k)$,we have $h=3$,$k=\frac{9-\lambda}{4}$,and $a=1$. The directrix of this parabola is $y = k+a$. Given the directrix is $y+1=0$,i.e.,$y=-1$,we equate $k+a = -1$. Substituting $k$ and $a$: $\frac{9-\lambda}{4} + 1 = -1$. $\frac{9-\lambda}{4} = -2 \implies 9-\lambda = -8 \implies \lambda = 17$. However,checking the options,if $\lambda=17$,then $k = \frac{9-17}{4} = -2$. The focus is $(h, k-a) = (3, -2-1) = (3, -3)$. Thus,option $C$ is correct.

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

Similar Questions

The equation of a tangent to the parabola $y^2 = 8x$ is $y = x + 2$. The point on this line from which the other tangent to the parabola is perpendicular to the given tangent is

If a chord of the parabola $y^2=4x$ passes through its focus and makes an angle $\theta$ with the $X$-axis,then its length is

Let $O$ be the origin and $A$ be a point on the curve $y^2=4x$. Then the locus of the midpoint of $OA$ is:

If $a > 0$ and $b^2 - 4ac = 0$,then the curve $y = ax^2 + bx + c$

The length of the latus rectum of the parabola $169\{(x-1)^2+(y-3)^2\}=(5x-12y+17)^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module