For the parabola y=x^2-3x+2,match the items i | vedclass

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

Question

(A) The given parabola is $y = x^2 - 3x + 2$. Rewriting it as $(x - \frac{3}{2})^2 = y + \frac{1}{4}$.Comparing with $(x-h)^2 = 4a(y-k)$,we get $h = \

Vedclass · Accepted Answer

$A-I, B-II, C-III, D-IV$

Vedclass · Answer

(A) The given parabola is $y = x^2 - 3x + 2$. Rewriting it as $(x - \frac{3}{2})^2 = y + \frac{1}{4}$.Comparing with $(x-h)^2 = 4a(y-k)$,we get $h = \frac{3}{2}$,$k = -\frac{1}{4}$,and $4a = 1 \implies a = \frac{1}{4}$.$1$. $Q$ is the vertex,which is $(h, k) = (\frac{3}{2}, -\frac{1}{4})$. Thus,$B-II$.$2$. $S$ is the focus $(h, k+a) = (\frac{3}{2}, -\frac{1}{4} + \frac{1}{4}) = (\frac{3}{2}, 0)$. Thus,$C-III$.$3$. $Z$ is the intersection of the axis $(x = \frac{3}{2})$ and directrix $(y = k-a = -\frac{1}{4} - \frac{1}{4} = -\frac{1}{2})$,so $Z = (\frac{3}{2}, -\frac{1}{2})$. Thus,$D-IV$.$4$. $P$ is an end point of the latus rectum $(h \pm 2a, k+a) = (\frac{3}{2} \pm \frac{1}{2}, 0)$. For $(2, 0)$,we have $A-I$.Therefore,the correct matching is $A-I, B-II, C-III, D-IV$.

$A$. $P$	$I$. $(2,0)$
$B$. $Q$	$II$. $(\frac{3}{2}, -\frac{1}{4})$
$C$. $S$	$III$. $(\frac{3}{2}, 0)$
$D$. $Z$	$IV$. $(\frac{3}{2}, -\frac{1}{2})$
	$V$. $(0, \frac{3}{2})$

Similar Questions

The normal drawn at a point $(2, -4)$ on the parabola $y^2 = 8x$ cuts the same parabola again at $(\alpha, \beta)$. Then $\alpha + \beta =$

If the normal at the point $(bt_1^2, 2bt_1)$ on the parabola $y^2 = 4bx$ meets the parabola again at the point $(bt_2^2, 2bt_2)$,then:

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:

Let $P$ be the point on the parabola ${y^2} = 8x$ which is at a minimum distance from the centre $C$ of the circle ${x^2} + {(y + 6)^2} = 1$. Then the equation of the circle,passing through $C$ and having its centre at $P$,is:

The angle between tangents to the parabola $y^2 = 4ax$ at the points where it intersects with the line $x - y - a = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module