Which of the following equations is the locus of $(at^2, 2at)$? (1) $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ (2) $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (3) $x^2+y^2=a^2$ (4) $y^2=4ax$

Which of the following equations is the locus of (at², 2at)? (1) (x²)/(a²)-(y²)/(b²)=1 (2) (x²)/(a²)+(y²)/(b²)

Exercise 6.5 · Q2

(at^2, 2at)

(1)

\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

(2)

\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1

(3)

x^2+y^2=a^2

(4)

y^2=4ax

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Concept understanding — Locus of a Point

A point is not a thing but a place — a position located, with reference to a coordinate system, by a unique ordered pair $(x,y)$ : $x$ is the (signed) distance from the $y$ -axis and $y$ the (signed) distance from the $x$ -axis. The locus of a moving point is the path traced out by it as it moves subject to a stated geometric condition; equivalently, an equation in $x,y$ has infinitely many real solution-pairs $(x,y)$ , and the collection of all the points corresponding to those solutions is the locus of the equation. (The plural of locus is loci.)

Some standard loci: a point moving so as to stay equidistant from two fixed points $A,B$ traces the perpendicular bisector of $AB$ ; a point equidistant from two fixed lines traces the angle bisector; a point at a fixed distance $r$ from a fixed point $O$ traces a circle of radius $r$ centred at $O$ . A point on the rim of a circle rolling along a straight line traces a cycloid.

Procedure for finding the equation of a locus.

Assign coordinates $(h,k)$ to the moving point $P$ whose locus is wanted.
Translate the given geometric condition(s) into equation(s) relating $h,k$ and any parameters/known quantities.
Eliminate the parameter(s), so that only $h,k$ and known constants remain.
Replace $h$ by $x$ and $k$ by $y$ ; the resulting equation in $x,y$ is the equation of the locus.

Tip

When the given coordinates involve a trigonometric parameter $\theta$ (e.g. $(a\sec\theta,\,b\tan\theta)$ ), eliminate $\theta$ using a Pythagorean identity: $\sin^2\theta+\cos^2\theta=1$ , $\sec^2\theta-\tan^2\theta=1$ , or $\csc^2\theta-\cot^2\theta=1$ . When the parameter is a real number $t$ (e.g. $(ct,\,c/t)$ ), eliminate it by taking the product, ratio, or another algebraic combination of the two coordinate equations that cancels $t$ .

Worked pattern. For $P$ equidistant from two fixed points $A(x_1,y_1)$ and $B(x_2,y_2)$ : set $P=(h,k)$ , write $PA=PB$ , square both sides, and expand — the $h^2,k^2$ terms cancel, leaving a linear equation (the perpendicular bisector). For $P$ at a fixed distance from a fixed point, $PA=r$ squared gives the circle equation directly. For a parametric locus $(f(\theta),g(\theta))$ , solve each coordinate equation for the trig function of $\theta$ it contains, then combine using an identity to eliminate $\theta$ entirely — this is exactly how the standard hyperbola/ellipse/astroid forms arise from their parametric descriptions.

The phrase "locus of a point class 11 maths definition and examples" shows up often in student searches tied to the Straight Lines chapter of the NCERT/CBSE Class 11 Mathematics curriculum, where locus problems are a recurring important-questions category in both board exams and JEE Main. Learning the standard four-step procedure here also pays off directly in later chapters on circles and conic sections that build on the same idea.

Eliminate the parameter $t$ from $x=at^2,\ y=2at$ using $t=\dfrac{y}{2a}$ .

The point $(at^2,2at)$ is given in parametric form; the locus is found by eliminating $t$ .

Step 1. Isolate $t$ from the simpler coordinate. From $y=2at$ , we get

t=\frac{y}{2a}.

Step 2. Substitute into $x=at^2$ .

x = a\left(\frac{y}{2a}\right)^2 = a\cdot\frac{y^2}{4a^2}=\frac{y^2}{4a}.

Step 3. Rearrange. Multiply both sides by $4a$ :

y^2 = 4ax.

This is the standard equation of a parabola with vertex at the origin opening along the positive $x$ -axis, and it is option (4). Options (1) and (2) are the standard hyperbola/ellipse forms, and option (3) is a circle — none of these arise from eliminating $t$ here.

✓Final answer

y^2=4ax

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

CBSE 2026Set ANNUAL1 markMCQ
$(a\cos\theta, b\sin\theta)$
(a) $x^2+y^2=a^2$
(b) $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$
(c) $y^2=4ax$
(d) $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
›Reveal solutionSolution

With $x=a\cos\theta$ , $y=b\sin\theta$ , using $\cos^2\theta+\sin^2\theta=1$ eliminates $\theta$ to give the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ .

Let $x=a\cos\theta$ and $y=b\sin\theta$ . Then $\cos\theta=\dfrac{x}{a}$ and $\sin\theta=\dfrac{y}{b}$ .

Using the Pythagorean identity $\cos^2\theta+\sin^2\theta=1$ :

$\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2=1 \;\Rightarrow\; \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$

This is the standard equation of an ellipse — the point $(a\cos\theta,b\sin\theta)$ is the well-known parametric form of a point on an ellipse.

✓Final answer
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$
CBSE 2024Set ANNUAL1 markMCQ
$(8,-5)$
(a) $2$
(b) $0$
(c) $3$
(d) $1$
›Reveal solutionSolution

Substituting $(8,-5)$ into the locus equation gives $k=3$ .

The locus is $\dfrac{x^2}{16}-\dfrac{y^2}{25}=k$ . Since $(8,-5)$ lies on it, put $x=8,\ y=-5$ :

$\dfrac{8^2}{16}-\dfrac{(-5)^2}{25}=\dfrac{64}{16}-\dfrac{25}{25}=4-1=3.$

So $k=3$ .

✓Final answer
$k=3$
CBSE 2023Set ANNUAL1 markMCQ
$3x^2 + 3y^2 - 8x - 12y + 17 = 0$
(a) (1, 2)
(b) (0, 0)
(c) (0, -1)
(d) (-2, 3)
›Reveal solutionSolution

Substitute each option into the equation; the point that makes it exactly zero lies on the curve.

Test $(1,2)$ : $3(1)^2+3(2)^2-8(1)-12(2)+17 = 3+12-8-24+17 = 0$ . ✓ It satisfies the equation.

Test $(0,0)$ : $0+0-0-0+17=17\ne 0$ . ✗

Test $(0,-1)$ : $0+3-0+12+17=32\ne 0$ . ✗

Test $(-2,3)$ : $12+27+16-36+17=36\ne 0$ . ✗

Only $(1,2)$ lies on the locus.

✓Final answer
$(1,2)$
CBSE 2020Set ANNUAL1 markMCQ
$(8, -5)$
(a) 0
(b) 1
(c) 2
(d) 3
›Reveal solutionSolution

Direct substitution of the point $(8,-5)$ into the locus equation gives $k=3$ .

A point lies on a locus if its coordinates satisfy the locus equation. Substituting $x=8$ , $y=-5$ :

$\dfrac{(8)^2}{16}-\dfrac{(-5)^2}{25}=\dfrac{64}{16}-\dfrac{25}{25}=4-1=3.$

So $k=3$ .

✓Final answer
$The correct option is (d) 3.$

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Q.Which of the following equations is the locus of (at2,2at)(at^2, 2at)(at2,2at)?

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Q.Which of the following equations is the locus of $(at^2, 2at)$ ?

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