If the vertex of a parabola is (4,3) and its | vedclass

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

Question

(B) The vertex $V$ is $(4,3)$ and the directrix $L$ is $3x+2y-7=0$. The axis of the parabola is perpendicular to the directrix and passes through the

Vedclass · Accepted Answer

$3x+2y-29=0$

Vedclass · Answer

(B) The vertex $V$ is $(4,3)$ and the directrix $L$ is $3x+2y-7=0$. The axis of the parabola is perpendicular to the directrix and passes through the vertex. The slope of the directrix is $m = -3/2$. Thus,the slope of the axis is $m' = 2/3$. The equation of the axis is $y-3 = \frac{2}{3}(x-4)$,which simplifies to $2x-3y+1=0$. The focus $S$ lies on the axis at a distance $a$ from the vertex,where $a$ is the distance from the vertex to the directrix. The distance $a = \frac{|3(4)+2(3)-7|}{\sqrt{3^2+2^2}} = \frac{|12+6-7|}{\sqrt{13}} = \frac{11}{\sqrt{13}}$. The latus rectum is parallel to the directrix and passes through the focus. The distance from the vertex to the latus rectum is $a$. The equation of the latus rectum is $3x+2y+k=0$. Since the vertex is at distance $a$ from the directrix and the focus is at distance $2a$ from the directrix,the latus rectum is at distance $2a$ from the directrix. The equation of the latus rectum is $3x+2y+k=0$. The distance between $3x+2y-7=0$ and $3x+2y+k=0$ is $2a = \frac{22}{\sqrt{13}}$. $\frac{|k-(-7)|}{\sqrt{13}} = \frac{22}{\sqrt{13}} \implies |k+7| = 22$. $k+7 = 22 \implies k=15$ or $k+7 = -22 \implies k=-29$. Since the focus must be on the opposite side of the vertex from the directrix,we find the focus $S = (4+2a\cos	heta, 3+2a\sin	heta)$ where $\cos	heta = \frac{3}{\sqrt{13}}, \sin	heta = \frac{2}{\sqrt{13}}$. $S = (4 + 2(\frac{11}{\sqrt{13}})(\frac{3}{\sqrt{13}}), 3 + 2(\frac{11}{\sqrt{13}})(\frac{2}{\sqrt{13}})) = (4 + \frac{66}{13}, 3 + \frac{44}{13}) = (\frac{118}{13}, \frac{83}{13})$. Substituting $S$ into $3x+2y+k=0$: $3(\frac{118}{13}) + 2(\frac{83}{13}) + k = 0 \implies \frac{354+166}{13} + k = 0 \implies k = -\frac{520}{13} = -40$. Wait,checking the distance again: The latus rectum is $3x+2y+k=0$. The vertex $(4,3)$ is at distance $a$ from directrix. The latus rectum is at distance $2a$ from directrix. The equation is $3x+2y-29=0$.

Column-$I$	Column-$II$
$(A)$ Circumradius of $\Delta PQR$	$(P)$ $5/2$
$(B)$ Area of $\Delta PQR$	$(Q)$ $(5/2, 0)$
$(C)$ Centroid of $\Delta PQR$	$(R)$ $(2/3, 0)$
$(D)$ Circumcenter of $\Delta PQR$	$(S)$ $2$

If the vertex of a parabola is $(4,3)$ and its directrix is $3x+2y-7=0$,then the equation of the latus rectum of the parabola is

Similar Questions

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

Let $PQ$ be a double ordinate of the parabola $y^2 = -4x$,where $P$ lies in the second quadrant. If $R$ divides $PQ$ in the ratio $2 : 1$,then the locus of $R$ is:

The locus of the point of intersection of the tangents drawn at the ends of a focal chord of the parabola $x^{2}=-8y$ is

$P$ and $Q$ are two distinct points on the parabola,$y^2 = 4x$,with parameters $t$ and $t_1$ respectively. If the normal at $P$ passes through $Q$,then the minimum value of $t_1^2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module