If $P(2, -7)$ is a given point and $Q$ is a point on $2x^2 + 9y^2 = 18$, then find the equation of the locus of the mid-point of $PQ$.

If P(2, -7) is a given point and Q is a point on 2x² + 9y² = 18, then find the equation of the locus of the mi

Exercise 6.1 · Q10

P(2, -7)

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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 $t$ . …

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Express $Q$ 's coordinates in terms of the midpoint of $PQ$ and substitute into the ellipse equation $Q$ satisfies.

Step 1. Set up. Let $M=(x,y)$ be the midpoint of $PQ$ , where $P=(2,-7)$ and $Q=(x_Q,y_Q)$ lies on $2x^2+9y^2=18$ . By the midpoint formula,

x=\frac{2+x_Q}{2}, \qquad y=\frac{-7+y_Q}{2}.

Step 2. Solve for $x_Q,y_Q$ .

x_Q=2x-2, \qquad y_Q=2y+7.

Step 3. Substitute into the ellipse equation. Since $2x_Q^2+9y_Q^2=18$ :

2(2x-2)^2+9(2y+7)^2=18.

Step 4. Expand $(2x-2)^2$ and $(2y+7)^2$ .

$(2x-2)^2=4x^2-8x+4, \qquad (2y+7)^2=4y^2+28y+49.$ …

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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 $\cos^2\theta+\sin^2\theta=1$ :

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

…

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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 $x=8,\ y=-5$ : …

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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 $(0,0)$ : $0+0-0-0+17=17\ne 0$ . ✗

…

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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 $x=8$ , $y=-5$ : …

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Q.If P(2,−7)P(2, -7)P(2,−7) is a given point and QQQ is a point on 2x2+9y2=182x^2 + 9y^2 = 182x2+9y2=18, then find the equation of the locus of the mid-point of PQPQPQ.

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Q.If $P(2, -7)$ is a given point and $Q$ is a point on $2x^2 + 9y^2 = 18$ , then find the equation of the locus of the mid-point of $PQ$ .

Unlock everything free for 14 days