(A) The given equation is $y^{2}=8x$.Comparing this with the standard equation of a parabola $y^{2}=4ax$,we get $4a=8$,which implies $a=2$.Since the e

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Focus: $(2, 0)$,Axis: $y=0$,Directrix: $x=-2$,Latus Rectum: $8$

(A) The given equation is $y^{2}=8x$.Comparing this with the standard equation of a parabola $y^{2}=4ax$,we get $4a=8$,which implies $a=2$.Since the equation is of the form $y^{2}=4ax$,the axis of symmetry is the $x$-axis (i.e.,$y=0$).The focus of the parabola is $(a, 0) = (2, 0)$.The equation of the directrix is $x=-a$,which is $x=-2$.The length of the latus rectum is $4a = 4 \times 2 = 8$.

Class 11 Mathematics 2 Conic Sections: Parabola, Ellipse and Hyperbola Parabola

Easy

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

A
Focus: $(2, 0)$,Axis: $y=0$,Directrix: $x=-2$,Latus Rectum: $8$
B
Focus: $(0, 2)$,Axis: $x=0$,Directrix: $y=-2$,Latus Rectum: $8$
C
Focus: $(2, 0)$,Axis: $x=0$,Directrix: $x=2$,Latus Rectum: $4$
D
Focus: $(-2, 0)$,Axis: $y=0$,Directrix: $x=2$,Latus Rectum: $8$

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2 Conic Sections: Parabola, Ellipse and Hyperbola

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Parabola

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Let $P$ be the point $(1, 0)$ and $Q$ be a point on the locus $y^2 = 8x$. The locus of the midpoint of $PQ$ is

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Let $ABCD$ be a trapezium whose vertices lie on the parabola $y^2=4x$. Let the sides $AD$ and $BC$ of the trapezium be parallel to the $y$-axis. If the diagonal $AC$ is of length $\frac{25}{4}$ and it passes through the point $(1,0)$,then the area of $ABCD$ is:

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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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