Find the equation of the parabola with vertex (-1, -2), axis parallel to y-axis and passing through (3, 6).

Question 113 of 126

(-1, -2)

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023Subjective· 3mImportance★★★★★

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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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With axis parallel to the $y$ -axis, the parabola is $(x+1)^2=4a(y+2)$ ; substituting the given point fixes $a$ .

Vertex $(-1,-2)$ , axis parallel to the $y$ -axis $\Rightarrow$ equation of the form $(x-(-1))^2=4a(y-(-2))$ , i.e. $(x+1)^2=4a(y+2)$ . …

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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 equation of the parabola with vertex (−1,−2)(-1, -2)(−1,−2), axis parallel to yyy-axis and passing through (3,6)(3, 6)(3,6).

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Q.Find the equation of the parabola with vertex $(-1, -2)$ , axis parallel to $y$ -axis and passing through $(3, 6)$ .

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