The axis of a parabola is along the line y=x. | vedclass

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

Question

(A) The axis of the parabola is $y=x$. The vertex $A$ and focus $S$ lie on this line in the first quadrant. Since the distance of $A$ from $(0,0)$ is

Vedclass · Accepted Answer

$x=(t+1)^2, y=(t-1)^2$

Vedclass · Answer

(A) The axis of the parabola is $y=x$. The vertex $A$ and focus $S$ lie on this line in the first quadrant. Since the distance of $A$ from $(0,0)$ is $\sqrt{2}$,the coordinates of $A$ are $(1,1)$. Since the distance of $S$ from $(0,0)$ is $2\sqrt{2}$,the coordinates of $S$ are $(2,2)$. The distance $AS = a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$. The directrix is perpendicular to the axis $y=x$ and passes through a point $Z$ such that $A$ is the midpoint of $SZ$. Since $S=(2,2)$ and $A=(1,1)$,$Z=(0,0)$. The equation of the directrix is $x+y=0$. By the definition of a parabola,the distance from any point $(x,y)$ to the focus $(2,2)$ equals the distance to the directrix $x+y=0$: $(x-2)^2 + (y-2)^2 = \frac{(x+y)^2}{2}$. $2(x^2 - 4x + 4 + y^2 - 4y + 4) = x^2 + y^2 + 2xy$. $x^2 + y^2 - 2xy - 8x - 8y + 16 = 0$,which is $(x-y)^2 = 8(x+y-2)$. Substituting $x=(t+1)^2$ and $y=(t-1)^2$: $((t+1)^2 - (t-1)^2)^2 = (4t)^2 = 16t^2$. $8((t+1)^2 + (t-1)^2 - 2) = 8(t^2 + 2t + 1 + t^2 - 2t + 1 - 2) = 8(2t^2) = 16t^2$. Both sides match,so the parametric form is $x=(t+1)^2, y=(t-1)^2$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

The axis of a parabola is along the line $y=x$. The distance of its vertex $A$ from $(0,0)$ is $\sqrt{2}$ and that of its focus $S$ from $(0,0)$ is $2\sqrt{2}$. If $A$ and $S$ lie in the first quadrant,then the equation of the parabola in parametric form is

Similar Questions

The length of the latus rectum of the parabola $x^{2}+2y=8x-7$ is

If the line $y = -x + k$ is normal to the curve $y^2 = 16x$,then $k$ is

The locus of a point which divides the line segment joining the focus and any point on the parabola $y^2 = 12x$ in the ratio $m:n$ $(m+n \neq 0)$ is a parabola. Then the length of the latus rectum of that parabola is

Find the coordinates of the focus,axis of the parabola,the equation of the directrix,and the length of the latus rectum for $y^{2}=12x$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module