(B) Let $x = (u \cos \alpha)t$ and $y = (u \sin \alpha)t - \frac{1}{2}gt^2$.From the first equation,we have $t = \frac{x}{u \cos \alpha}$.Substituting

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(B) Let $x = (u \cos \alpha)t$ and $y = (u \sin \alpha)t - \frac{1}{2}gt^2$.From the first equation,we have $t = \frac{x}{u \cos \alpha}$.Substituting this value of $t$ into the equation for $y$:$y = (u \sin \alpha) \left( \frac{x}{u \cos \alpha} \right) - \frac{1}{2}g \left( \frac{x}{u \cos \alpha} \right)^2$$y = x \tan \alpha - \frac{g x^2}{2 u^2 \cos^2 \alpha}$.This equation is of the form $y = ax^2 + bx + c$,which represents a parabola.

Class 11 Mathematics 10-2. Parabola, Ellipse, Hyperbola Parabola

Medium

The position of a moving point in the $XY$-plane at time $t$ is given by $\left( (u \cos \alpha)t, (u \sin \alpha)t - \frac{1}{2}gt^2 \right)$,where $u, \alpha, g$ are constants. The locus of the moving point is

A
$A$ circle
B
$A$ parabola
C
An ellipse
D
None of these

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Class 11

Mathematics Questions

Chapter

10-2. Parabola, Ellipse, Hyperbola

Subtopic

Parabola

From the point $(-1, -6)$, two tangents are drawn to the parabola $y^2 = 4x$. The angle between the two tangents is:

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If the parabola $x^2=4ay, (a>0)$ makes an intercept of length $\sqrt{40}$ units on the line $y=1+2x$,then $4a=$

MediumTS EAMCET 2020

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For the parabola $y^2+6y-2x+5=0$,match the items in List-$I$ with the suitable item in List-$II$ given below:

List-$I$ (Geometric Property) List-$II$ (Coordinates/Equations)

$I$. Vertex $A$. $\left(-\frac{3}{2}, -3\right)$

$II$. Focus $B$. $\left(\frac{3}{2}, -3\right)$

$III$. Equation of the directrix $C$. $2x + 5 = 0$

$IV$. Equation of the axis $D$. $2x + y + 3 = 0$

$E$. $y + 3 = 0$

$F$. $(-2, -3)$

The correct matching is:

List-$I$ (Geometric Property)	List-$II$ (Coordinates/Equations)
$I$. Vertex	$A$. $\left(-\frac{3}{2}, -3\right)$
$II$. Focus	$B$. $\left(\frac{3}{2}, -3\right)$
$III$. Equation of the directrix	$C$. $2x + 5 = 0$
$IV$. Equation of the axis	$D$. $2x + y + 3 = 0$
	$E$. $y + 3 = 0$
	$F$. $(-2, -3)$

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Let the point $P$ of the focal chord $PQ$ of the parabola $y^2=16x$ be $(1, -4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$,where $\operatorname{gcd}(m, n)=1$,then $m^2+n^2$ is equal to:

DifficultJEE MAIN 2025

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The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

MediumAP EAMCET 2017

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The position of a moving point in the $XY$-plane at time $t$ is given by $\left( (u \cos \alpha)t, (u \sin \alpha)t - \frac{1}{2}gt^2 \right)$,where $u, \alpha, g$ are constants. The locus of the moving point is

Similar Questions

From the point $(-1, -6)$, two tangents are drawn to the parabola $y^2 = 4x$. The angle between the two tangents is:

If the parabola $x^2=4ay, (a>0)$ makes an intercept of length $\sqrt{40}$ units on the line $y=1+2x$,then $4a=$

Let the point $P$ of the focal chord $PQ$ of the parabola $y^2=16x$ be $(1, -4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$,where $\operatorname{gcd}(m, n)=1$,then $m^2+n^2$ is equal to:

The tangents to the parabola $y^2 = 4ax$ from an external point $P$ make angles $\theta_1$ and $\theta_2$ with the axis of the parabola,such that $\tan \theta_1 + \tan \theta_2 = b$,where $b$ is a constant. Then $P$ lies on

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