Let the parabola y = x^2 + px + q passing thr | vedclass

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

Question

(B) The parabola is given by $y = x^2 + px + q$. Since it passes through $(1, -1)$, we have $-1 = 1 + p + q$, which implies $q = -p - 2$.The vertex of

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) The parabola is given by $y = x^2 + px + q$. Since it passes through $(1, -1)$, we have $-1 = 1 + p + q$, which implies $q = -p - 2$.The vertex of the parabola $y = ax^2 + bx + c$ is at $(-b/2a, -D/4a)$. For $y = x^2 + px + q$, the vertex is $(-p/2, q - p^2/4)$.The distance from the vertex to the $x$-axis is the absolute value of the $y$-coordinate of the vertex, which is $d = |q - p^2/4|$.Substituting $q = -p - 2$, we get $d = |-p - 2 - p^2/4| = |p^2/4 + p + 2|$.To minimize $d$, we analyze the quadratic $f(p) = p^2/4 + p + 2$. The derivative $f'(p) = p/2 + 1$. Setting $f'(p) = 0$ gives $p = -2$.At $p = -2$, $q = -(-2) - 2 = 0$.The value of $p^2 + q^2 = (-2)^2 + 0^2 = 4 + 0 = 4$.

List-$I$	List-$II$
$(A)$ Focus	$(I)$ $(4,2)$
$(B)$ Vertex	$(II)$ $(3,2)$
$(C)$ One end of the latus rectum	$(III)$ $(2,3)$
$(D)$ Point of intersection of the axis and directrix	$(IV)$ $(2,4)$
	$(V)$ $(2,2)$

Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is:

Similar Questions

The angle between the tangents drawn from the point $(1,4)$ to the parabola $y^2=4x$ is

The shortest distance between the line $y-x=1$ and the curve $x=y^2$ is

Let $(x, y)$ be any point on the parabola $y^2 = 4x$. Let $P$ be the point that divides the line segment from $(0, 0)$ to $(x, y)$ in the ratio $1:3$. Then the locus of $P$ is

$A$ point on the parabola $y^2 = 18x$ at which the ordinate increases at twice the rate of the abscissa is

Vedclass Test Series

Exam Paper Generator

Online Exam Module