Find the equation of the parabola if the curve is open leftward, vertex is (2, 1) and passing through the poin

Question 103 of 126

Q. Find the equation of the parabola if the curve is open leftward, vertex is (2, 1) and passing through the point (1, 3).

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2020Subjective· 2mImportance★★★★★

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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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Sets up the standard leftward-opening parabola form at the given vertex, then solves for the parameter $a$ using the given point on the curve.

A parabola that opens leftward with vertex $(h,k)$ has the standard equation $(y-k)^2=-4a(x-h)$ , $a>0$ .
Here vertex $(h,k)=(2,1)$ , so the equation is $(y-1)^2=-4a(x-2)$ .
The curve passes through $(1,3)$ , so substitute $x=1,\ y=3$ : $(3-1)^2=-4a(1-2)$ .
Simplify: $4=-4a(-1)=4a$ , so $a=1$ . …

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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 if the curve is open leftward, vertex is (2, 1) and passing through the point (1, 3).

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