Find the axis, vertex, focus, equation of directrix, latus rectum, length of latus rectum for the parabola y²+

Question 75 of 126

y^2+8x-6y+1=0

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2016Subjective· 10mImportance★★★★★

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Concept understanding — Parabola — Standard Equation

Setting eccentricity $e=1$ in the conic definition ( $SP=e\cdot PM$ ) gives the parabola — the locus of points equidistant from a fixed focus and a fixed directrix.

Standard form (vertex at the origin, focus on the axis). With focus $(a,0)$ and directrix $x=-a$ : $y^2=4ax$ (opens right). The other three orientations, by symmetry: $y^2=-4ax$ (left), $x^2=4ay$ (up), $x^2=-4ay$ (down).

Vocabulary. Axis: line through the focus $\perp$ directrix. Vertex: where the axis meets the curve. Focal chord: any chord through the focus. Latus rectum: the focal chord $\perp$ the axis; for $y^2=4ax$ its endpoints are $(a,\pm2a)$ , so its length is $4a$ .

Vertex at $(h,k)$ :

Equation	Focus	Directrix
$(y-k)^2=4a(x-h)$	$(h+a,k)$	$x=h-a$
$(y-k)^2=-4a(x-h)$	$(h-a,k)$	$x=h+a$
$(x-h)^2=4a(y-k)$	$(h,k+a)$	$y=k-a$
$(x-h)^2=-4a(y-k)$	$(h,k-a)$	$y=k+a$

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Figure — Draw the left-opening parabola (y-3)^2 = -8(x-1)

Complete the square in $y$ to write the parabola in standard form $(y-k)^2=-4a(x-h)$ , then read off every required element.

1. Given equation.

$y^2+8x-6y+1=0$

2. Complete the square in $y$ .

$y^2-6y=-8x-1$

$y^2-6y+9=-8x-1+9$

$(y-3)^2=-8x+8=-8(x-1)$

3. Compare with the standard form $(y-k)^2=-4a(x-h)$ , which is a parabola opening in the $-x$ direction with vertex $(h,k)$ :

$4a=8 \ \Rightarrow\ a=2,\qquad (h,k)=(1,3)$

4. Vertex.

$(h,k)=(1,3)$

5. Axis of the parabola (the horizontal line through the vertex, parallel to the axis of symmetry):

$y=3$

6. Focus. Since the parabola opens toward $-x$ , the focus lies at $(h-a,k)$ :

$(1-2,\ 3)=(-1,3)$

7. Directrix. The directrix lies at $x=h+a$ :

$x=1+2=3$

8. Latus rectum. The latus rectum is the vertical line through the focus:

$x=-1$

and its length is: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set ANNUAL1 markMCQ
$x^2=24y$
(a) $8$
(b) $24$
(c) $6$
(d) $12$
›Reveal solutionSolution
Comparing $x^2=24y$ with the standard form $x^2=4ay$ directly gives the latus rectum length as $24$ .

The standard form of a parabola opening upward is $x^2=4ay$ , whose latus rectum has length $4a$ .

Comparing the given equation $x^2=24y$ with $x^2=4ay$ , we get $4a=24$ . …
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CBSE 2020Set ANNUAL1 markMCQ
$x^2 = 8y - 1$
(a) $\left(0, -\dfrac{1}{8}\right)$
(b) $\left(-\dfrac{1}{8}, 0\right)$
(c) $\left(\dfrac{1}{8}, 0\right)$
(d) $\left(0, \dfrac{1}{8}\right)$
›Reveal solutionSolution
Rewriting $x^2=8y-1$ in the standard form $x^2=4a(y-k)$ shows the vertex is $\left(0,\dfrac{1}{8}\right)$ .

Start with $x^2=8y-1$ .

Add $1$ to both sides and factor the right side: $x^2 = 8y-1 = 8\left(y-\dfrac{1}{8}\right)$ .

This is now in the standard vertical-axis parabola form $x^2=4a(y-k)$ , where the vertex is $(h,k)=(0,k)$ (here $h=0$ since there is no $x$ -shift, only a coefficient of $x^2$ ). …
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CBSE 2016Set ANNUAL1 markMCQ
$(2,-3)$
(a) $2$
(b) $4$
(c) $6$
(d) $8$
›Reveal solutionSolution
The length of the latus rectum is $4a=8$ , where $a$ is the vertex-to-directrix distance.

For a parabola, the vertex lies midway between the focus and the directrix, at distance $a$ from each.

Vertex $=(2,-3)$ , directrix $x=4$ (a vertical line), so the axis of the parabola is horizontal, $y=-3$ .

Distance from vertex to directrix: $a=|4-2|=2$ .

Length of latus rectum $=4a=4(2)=8$ . …
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Q.Find the axis, vertex, focus, equation of directrix, latus rectum, length of latus rectum for the parabola y2+8x−6y+1=0y^2+8x-6y+1=0y2+8x−6y+1=0 and also draw the diagram.

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Q.Find the axis, vertex, focus, equation of directrix, latus rectum, length of latus rectum for the parabola $y^2+8x-6y+1=0$ and also draw the diagram.

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