(A) Since the vertex is $(0, 0)$ and the axis of the parabola is the $x$-axis,the equation of the parabola is of the form $y^2 = 4ax$.The parabola pas

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(A) Since the vertex is $(0, 0)$ and the axis of the parabola is the $x$-axis,the equation of the parabola is of the form $y^2 = 4ax$.The parabola passes through the point $(2, 3)$,which lies in the first quadrant.Substituting the point $(2, 3)$ into the equation $y^2 = 4ax$:$3^2 = 4a(2)$$9 = 8a$$a = \frac{9}{8}$Thus,the equation of the parabola is:$y^2 = 4 \left( \frac{9}{8} \right) x$$y^2 = \frac{9}{2} x$$2y^2 = 9x$

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

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Find the equation of the parabola that satisfies the following conditions: Vertex $(0, 0)$,passing through $(2, 3)$,and the axis is along the $x$-axis.

A
$2y^2 = 9x$
B
$3y^2 = 4x$
C
$9y^2 = 2x$
D
$4y^2 = 9x$

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10-2. Parabola, Ellipse, Hyperbola

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Parabola

Let the locus of the points from which the tangents drawn to $y = x^2$ make an angle of $45^{\circ}$ with each other be $16y^2 - 16x^2 + ky + 1 = 0$. Then $k$ is equal to:

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If $y = mx + c$ is the normal at a point on the parabola $y^2 = 8x$ whose focal distance is $8 \text{ units}$, then $|c|$ is equal to (in $\sqrt{3}$)

DifficultJEE MAIN 2017

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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:

MediumAIEEE 2003

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Find the coordinates of the focus,axis,the equation of the directrix,and the length of the latus rectum of the parabola $y^{2}=8x$.

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The distance of the point $(6, -2 \sqrt{2})$ from the common tangent $y = mx + c$ $(m > 0)$ of the curves $x = 2y^2$ and $x = 1 + y^2$ is

DifficultJEE MAIN 2023

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Find the equation of the parabola that satisfies the following conditions: Vertex $(0, 0)$,passing through $(2, 3)$,and the axis is along the $x$-axis.

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Let the locus of the points from which the tangents drawn to $y = x^2$ make an angle of $45^{\circ}$ with each other be $16y^2 - 16x^2 + ky + 1 = 0$. Then $k$ is equal to:

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