Q.Find the equation of the set of points $P$, the sum of whose distances from $A(4, 0, 0)$ and $B(-4, 0, 0)$ is equal to $10$.

Showing the 12 most recent of 50 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Let $A(1,2,0)$ and $B(0,-2,-1)$ be two points. If $P$ is a variable point such that $PA = 2PB$ and the locus of $P$ is $a(x^2 + y^2 + z^2) + bx + cy + dz + 15 = 0$, then $2a + 3b + d =$ (A) $2c$ (B) $3c$ (C) $c$ (D) $4c$

   ›Reveal solutionSolution

   Expanding the distance condition $PA=2PB$ directly gives the locus equation, whose coefficients satisfy $2a+3b+d=c$.

   Concept and Intuition

   The locus of a point whose distances from two fixed points are in a constant ratio (Apollonius-type locus) is always a sphere in 3D. Squaring the ratio condition and expanding removes the square roots, directly yielding the sphere's equation.

   Step-by-Step Solution

   1. $PA^2=(x-1)^2+(y-2)^2+z^2 = x^2+y^2+z^2-2x-4y+5$.
   2. $PB^2=x^2+(y+2)^2+(z+1)^2 = x^2+y^2+z^2+4y+4+2z+1 = x^2+y^2+z^2+4y+2z+5$. (Recheck: $(y+2)^2=y^2+4y+4$, $(z+1)^2=z^2+2z+1$; total constant $4+1=5$.)
   3. $PA=2PB\Rightarrow PA^2=4PB^2$: $x^2+y^2+z^2-2x-4y+5 = 4(x^2+y^2+z^2+4y+2z+5)$.
   4. $x^2+y^2+z^2-2x-4y+5 = 4x^2+4y^2+4z^2+16y+8z+20$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.O is the origin. P is a point moving on the straight line $lx + my + n = 0$ ($n \neq 0$). If Q is a point on the segment $\overline{OP}$, such that $OP \cdot OQ = K^2$ where $K \neq 0$, Then the locus of Q is (A) $n(x^2 + y^2) = K^2(lx + my)$ (B) $K^2(x^2 + y^2) = n(lx + my)$ (C) $n(x^2 - y^2) = K^2(lx - my)$ (D) $n^2(x^2 + y^2) = K^2(lx - my)$

   ›Reveal solutionSolution

   Parametrize $P$ on the line, express $Q$ as a scaled copy of $P$ along the same ray using the given product-of-distances condition, then substitute back into the line's equation to eliminate $P$'s coordinates. The resulting locus of $Q$ is a circle through the origin: option (A).

   Concept and Intuition

   This is a classic "inverse point" locus problem. $Q$ is obtained from $P$ by scaling along the same ray from $O$ so that the product of the two distances $OP$ and $OQ$ is the constant $K^2$ — this is exactly the geometric definition of inversion in a circle of radius $K$ centered at $O$. Since $P$ is restricted to a straight line not through the origin, its inverse point $Q$ traces a circle passing through the origin (a standard result: inversion maps a line not through the center to a circle through the center).

   Step-by-Step Solution

   1. Let $P=(h,k)$ lie on the given line: $lh+mk+n=0$.
   2. Since $Q$ lies on segment $OP$ (same ray from $O$ as $P$), write $Q=(x,y)=tP=(th,tk)$ for some scalar $t>0$.
   3. The condition $OP\cdot OQ = K^2$ (product of the two distances) gives:

   $|OP|\cdot|OQ| = K^2 \implies \sqrt{h^2+k^2}\cdot t\sqrt{h^2+k^2}=K^2 \implies t(h^2+k^2)=K^2.$

   4. So $t = \dfrac{K^2}{h^2+k^2}$, and hence $x=th=\dfrac{K^2h}{h^2+k^2}$, $y=tk=\dfrac{K^2k}{h^2+k^2}$.
   5. Compute $x^2+y^2 = t^2(h^2+k^2) = t\cdot\big[t(h^2+k^2)\big] = t\cdot K^2$, so $t=\dfrac{x^2+y^2}{K^2}$.
   6. From $x=th$: $h=\dfrac{x}{t}=\dfrac{xK^2}{x^2+y^2}$. Similarly $k=\dfrac{yK^2}{x^2+y^2}$.
   7. Substitute into the line's equation $lh+mk+n=0$:

   $l\cdot\frac{xK^2}{x^2+y^2}+m\cdot\frac{yK^2}{x^2+y^2}+n=0.$

   8. Multiply through by $(x^2+y^2)$ and rearrange to isolate the constant term: …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $P(3,4)$ is a fixed point and $Q$ is a variable point on the circle $x^2+y^2=16$, then the locus of the midpoint of $PQ$ is (A) $x^2+y^2-3x-4y+\dfrac{9}{4}=0$ (B) $x^2+y^2+3x+4y+16=0$ (C) $x^2+y^2-3x+4y+9=0$ (D) $x^2+y^2+3x-4y+\dfrac{9}{4}=0$

   ›Reveal solutionSolution

   Express $Q$ in terms of the midpoint $M$ and substitute into the circle equation to get the locus, which is itself a circle.

   Concept and Intuition

   The midpoint locus problem is solved by parametrizing the moving point ($Q$) in terms of the locus variable (the midpoint $M$), since $P$ is fixed. If $M=(x,y)$ is the midpoint of $PQ$, then $Q=2M-P$. Substituting this into the known equation of the circle that $Q$ traces gives the equation satisfied by $M$ — the required locus.

   Step-by-Step Solution

   1. Let $M(x,y)$ be the midpoint of $P(3,4)$ and $Q(x_1,y_1)$ on $x^2+y^2=16$.
   2. $x=\dfrac{3+x_1}{2},\ y=\dfrac{4+y_1}{2}\Rightarrow x_1=2x-3,\ y_1=2y-4$.
   3. Since $Q$ lies on the circle: $x_1^2+y_1^2=16\Rightarrow (2x-3)^2+(2y-4)^2=16$. …

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4. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Consider the lines $L_1: 2x + 3y + 1 = 0$ and $L_2: 3x - 2y + 1 = 0$. The locus of a variable point that is equidistant from the two lines $L_1 = 0$ and $L_2 = 0$ is (A) $5x^2 - 24xy - 5y^2 + 2x - 10y = 0$ (B) $x - 5y = 0$ (C) $5x + y = 0$ (D) $5x^2 - 24xy - 5y^2 = 0$

   ›Reveal solutionSolution

   The locus equidistant from two lines is the pair of angle bisectors, combined into a single second-degree equation: $5x^2-24xy-5y^2+2x-10y=0$.

   Concept and Intuition

   A point equidistant from two lines $L_1=0$ and $L_2=0$ lies on one of their two angle bisectors. The combined (both bisectors together) locus is obtained by squaring the equidistance condition $\frac{L_1}{|n_1|}=\pm\frac{L_2}{|n_2|}$, giving a single quadratic equation (a degenerate conic — pair of straight lines) when $|n_1|=|n_2|$.

   Step-by-Step Solution

   1. $L_1: 2x+3y+1=0$, normal magnitude $\sqrt{2^2+3^2}=\sqrt{13}$. $L_2: 3x-2y+1=0$, normal magnitude $\sqrt{3^2+2^2}=\sqrt{13}$. Equal magnitudes, and $2\cdot3+3\cdot(-2)=0$ confirms $L_1\perp L_2$.
   2. Equidistance: $\dfrac{|2x+3y+1|}{\sqrt{13}}=\dfrac{|3x-2y+1|}{\sqrt{13}} \Rightarrow (2x+3y+1)^2=(3x-2y+1)^2$.
   3. Factor as a difference of squares: $[(2x+3y+1)-(3x-2y+1)]\,[(2x+3y+1)+(3x-2y+1)]=0$ $\Rightarrow (-x+5y)(5x+y+2)=0 \Rightarrow (x-5y)(5x+y+2)=0$.
   4. Expand: $(x-5y)(5x+y+2) = 5x^2+xy+2x-25xy-5y^2-10y = 5x^2-24xy-5y^2+2x-10y$. …

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5. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.The locus of $|z - 2i| + |z + 4i| = 10$ is (A) a circle with centre at $(0, -1)$ and radius $5$ (B) a parabola with focus at $(0, -1)$ (C) a hyperbola with foci at $(0, 2)$ and $(0, -4)$ (D) an ellipse with eccentricity $3/5$

   ›Reveal solutionSolution

   The equation $|z - 2i| + |z + 4i| = 10$ describes the set of points whose sum of distances to two fixed points is constant — that is the definition of an ellipse. The foci are at $(0,2)$ and $(0,-4)$, the center is $(0,-1)$, the major axis length is $10$, and the eccentricity is $3/5$, so the correct option is (D).

   The key is to recognize the geometric meaning of the equation. For a complex number $z = x + iy$, the expression $|z - a|$ is the distance from the point $(x,y)$ to the point representing $a$ in the complex plane. So $|z - 2i|$ is the distance from $(x,y)$ to $(0,2)$, and $|z + 4i|$ is the distance from $(x,y)$ to $(0,-4)$. The equation says: the sum of these two distances is constant (10). That is exactly the definition of an ellipse: the set of all points where the sum of distances to two fixed points (the foci) is constant.

   1. Identify the foci.

      The two fixed points are $(0,2)$ and $(0,-4)$. These are the foci of the ellipse. Notice they lie on the vertical line $x=0$.

   2. Find the center.

      The center of an ellipse is the midpoint of the segment joining the foci.

      Midpoint: $\left( \frac{0+0}{2}, \frac{2 + (-4)}{2} \right) = (0, -1)$.

   3. Determine the major axis length.

      The constant sum $10$ is the length of the major axis, so $2a = 10$, hence $a = 5$.

   4. Find the distance between the foci.

      Distance between $(0,2)$ and $(0,-4)$ is $6$, so $2c = 6$, hence $c = 3$.

   5. Compute the eccentricity.

      For an ellipse, eccentricity $e = \frac{c}{a} = \frac{3}{5}$.

   6. Check the options. …

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6. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If a tangent drawn to the parabola $y^2 = 16x$ meets the curve $xy = 4$ at the points $P$ and $Q$, then the locus of midpoint of $PQ$ is (A) $y^2 = 2x$ (B) $y^2 + 2x = 0$ (C) $y^2 = 4x$ (D) $y^2 + 4x = 0$

   ›Reveal solutionSolution

   This tests writing the tangent to a parabola in parametric form, intersecting it with a rectangular hyperbola to get a quadratic whose roots are the intersection points' coordinates, and eliminating the parameter to find the midpoint's locus. The locus is $y^2+2x=0$.

   Concept and Intuition

   The tangent to $y^2=4ax$ at the point with parameter $t$ (i.e. at $(at^2,2at)$) has the compact equation $ty=x+at^2$. Substituting this line into a different curve's equation ($xy=4$) produces a quadratic whose two roots are exactly the $y$-coordinates of the two intersection points $P,Q$. Vieta's formulas then give the sum and product of $y_1,y_2$ directly in terms of the tangent's parameter $t$, without solving for $P,Q$ individually. To get the locus of the midpoint, we express the midpoint's coordinates $(X,Y)$ in terms of $t$ and then eliminate $t$.

   Step-by-Step Solution

   1. Parabola $y^2=16x\Rightarrow 4a=16\Rightarrow a=4$. Tangent at parameter $t$: $ty=x+4t^2$, i.e. $x=ty-4t^2$.
   2. Substitute into $xy=4$:

   $(ty-4t^2)\,y=4 \;\Rightarrow\; ty^2-4t^2y-4=0.$

   3. This is a quadratic in $y$ with roots $y_1,y_2$ (the $y$-coordinates of $P,Q$):

   $y_1+y_2=\frac{4t^2}{t}=4t,\qquad y_1y_2=\frac{-4}{t}.$

   4. Midpoint $y$-coordinate: $Y=\dfrac{y_1+y_2}{2}=2t$.
   5. Since $P,Q$ lie on $xy=4$, $x_1=4/y_1$, $x_2=4/y_2$. Midpoint $x$-coordinate: $X=\frac{x_1+x_2}{2}=\frac12\left(\frac4{y_1}+\frac4{y_2}\right)=2\cdot\frac{y_1+y_2}{y_1y_2}=2\cdot\frac{4t}{-4/t}=2\cdot(-t^2)=-2t^2.$ …

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7. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If the distance of a point $P$ on an ellipse from its focus $(1, 2)$ is half of the distance of $P$ from its corresponding directrix $x + y = 0$, then the point of intersection of the given directrix and its major axis, is (A) $(2, -2)$ (B) $\left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$ (C) $(-1, 1)$ (D) $\left(\dfrac{1}{3}, -\dfrac{1}{3}\right)$

   ›Reveal solutionSolution

   This tests the focus-directrix definition of a conic (the eccentricity ratio $e=1/2$ identifies it as an ellipse, though the actual value of $e$ isn't even needed) and the geometric fact that the major axis is the line through the focus perpendicular to the corresponding directrix. Their intersection is $\left(-\frac12,\frac12\right)$.

   Concept and Intuition

   For any conic, a point $P$ on it satisfies $\dfrac{\text{distance to focus}}{\text{distance to directrix}}=e$ (the eccentricity), by definition. Here we're told this ratio is $\frac12$ for every point $P$ — that number itself is a red herring for what's actually asked. What matters is a separate, purely geometric fact: the major axis of a conic is always perpendicular to the directrix and passes through the corresponding focus. So regardless of the eccentricity's value, we can find the required intersection point using only the focus and the directrix's equation.

   Step-by-Step Solution

   1. Focus $=(1,2)$, directrix: $x+y=0$.
   2. The directrix has direction vector $(1,-1)$ (along $x+y=0$, e.g. moving from $(0,0)$ to $(1,-1)$ stays on the line). Its normal direction is therefore $(1,1)$.
   3. The major axis passes through the focus and is perpendicular to the directrix, i.e. it points along the normal direction $(1,1)$. Parametrise the major axis as

   $(x,y)=(1+s,\,2+s),\quad s\in\mathbb R.$

   4. Find where this line meets the directrix $x+y=0$: $(1+s)+(2+s)=0\;\Rightarrow\;3+2s=0\;\Rightarrow\;s=-\frac32.$ …

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8. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$, then the locus of the midpoints of the intercepts made by the tangents between the coordinate axes is (A) $\dfrac{x^2}{4}+\dfrac{y^2}{2}=1$ (B) $\dfrac{x^2}{2}+\dfrac{y^2}{4}=1$ (C) $\dfrac{1}{4x^2}+\dfrac{1}{2y^2}=1$ (D) $\dfrac{1}{2x^2}+\dfrac{1}{4y^2}=1$

   ›Reveal solutionSolution

   Using the parametric tangent to the ellipse, find where it cuts the two axes, take the midpoint of that intercept-segment, and eliminate the parameter $\theta$ to get the locus.

   Concept and Intuition

   Every tangent to an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $(a\cos\theta,b\sin\theta)$ has the simple equation $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$. This line meets the axes at two easily computed points, and the midpoint of the segment between them depends only on $\theta$. Eliminating $\theta$ using the Pythagorean identity converts this parametric midpoint into a Cartesian locus equation.

   Step-by-Step Solution

   1. $x^2+2y^2=2 \Rightarrow \frac{x^2}{2}+\frac{y^2}{1}=1$, so $a^2=2$, $b^2=1$.
   2. Tangent at parameter $\theta$: $\frac{x\cos\theta}{\sqrt2}+\frac{y\sin\theta}{1}=1$ (general form $\frac{x\cos\theta}{a}+\frac{y\sin\theta}{b}=1$).
   3. X-intercept ($y=0$): $x=\frac{a}{\cos\theta}$. Y-intercept ($x=0$): $y=\frac{b}{\sin\theta}$.
   4. Midpoint of the intercepted segment: $(x,y)=\left(\frac{a}{2\cos\theta},\frac{b}{2\sin\theta}\right)$.
   5. So $\cos\theta=\frac{a}{2x}$ and $\sin\theta=\frac{b}{2y}$.
   6. Use $\sin^2\theta+\cos^2\theta=1$: $\frac{a^2}{4x^2}+\frac{b^2}{4y^2}=1$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The tangents drawn to the hyperbola $5x^2 - 9y^2 = 90$ through a variable point P make the angles $\alpha$ and $\beta$ with its transverse axis. If $\alpha$, $\beta$ are the complementary angles, then the locus of P is (A) $x^2 + y^2 = 8$ (B) $x^2 - y^2 = 8$ (C) $x^2 - y^2 = 28$ (D) $x^2 + y^2 = 28$

   ›Reveal solutionSolution

   Using $m_1m_2=1$ (from the complementary-angle condition) in the tangent-pair relation for the hyperbola gives the locus $x^2-y^2=a^2+b^2=28$.

   Concept and Intuition

   From any external point, two tangent lines can be drawn to a hyperbola; their slopes are the two roots of a quadratic obtained by substituting the general tangent condition. When a geometric condition is imposed on the slopes (here, that the two inclination angles are complementary), it becomes a condition on the roots of that quadratic — most simply the product of roots — which converts directly into an equation in the point's coordinates, i.e. its locus.

   Step-by-Step Solution

   1. Standard form: $5x^2-9y^2=90\Rightarrow \dfrac{x^2}{18}-\dfrac{y^2}{10}=1$, so $a^2=18,\ b^2=10$.
   2. A tangent of slope $m$ to this hyperbola: $y=mx\pm\sqrt{a^2m^2-b^2}$.
   3. For it to pass through $P(h,k)$: $(k-mh)^2=a^2m^2-b^2$, which expands to $m^2(h^2-a^2)-2hkm+(k^2+b^2)=0$.
   4. Product of the two slopes: $m_1m_2=\dfrac{k^2+b^2}{h^2-a^2}$. …

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10. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.A straight line passing through a fixed point (2, 3) intersects the coordinate axes at points P and Q. If O is the origin and R is a variable point such that OPRQ is a rectangle, then the locus of R is (A) $3x + 2y = xy$ (B) $2x + 3y = xy$ (C) $3x + 2y = 6$ (D) $3x + 2y = 6xy$

    ›Reveal solutionSolution

    Use the intercept form of the line through $(2,3)$ and the rectangle geometry ($R$ is the fourth vertex opposite $O$) to get the locus. Answer: $3x+2y=xy$.

    Concept and Intuition

    A line cutting the axes at $P=(a,0)$ and $Q=(0,b)$ has intercept form $\dfrac xa+\dfrac yb=1$. Since $OP$ lies along the x-axis and $OQ$ along the y-axis, they are perpendicular, so the quadrilateral $OPRQ$ (with $R$ diagonally opposite $O$) is automatically a rectangle, and $R$ must be the point $(a,b)$ — directly "completing" the rectangle. As the line varies (always through the fixed point $(2,3)$), $(a,b)$ traces the locus of $R$.

    Step-by-Step Solution

    1. Let the variable line meet the x-axis at $P(a,0)$ and the y-axis at $Q(0,b)$. Its equation is $\dfrac xa+\dfrac yb=1$.
    2. Since the line always passes through the fixed point $(2,3)$: $\frac2a+\frac3b=1 \quad (\ast)$ …

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11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.Let A(5,4) and B(5,-4) be two points. If P is a point in the coordinate plane such that $\angle APB = \dfrac{\pi}{4}$, then the point P lies on the curve (A) $x^2+y^2+10x-17=0$ (B) $x^2+y^2-2x-31=0$ (C) $x^2+y^2-10x+17=0$ (D) $x^2+y^2+2x-31=0$

    ›Reveal solutionSolution

    A fixed angle subtended by a fixed segment traces a circular arc (inscribed-angle theorem); working out the specific circle gives $x^2+y^2-2x-31=0$.

    Concept and Intuition

    When a segment $AB$ subtends a constant angle $\theta$ at a variable point $P$, $P$ traces an arc of a circle through... more precisely, two circular arcs (one on each side of $AB$) where $AB$ is a chord subtending inscribed angle $\theta$. The radius of such a circle is fixed by $AB = 2R\sin\theta$.

    Step-by-Step Solution

    1. $A=(5,4)$, $B=(5,-4)$: length $AB=8$, midpoint $M=(5,0)$, and $AB$ is vertical.
    2. By the inscribed angle / chord relation, $AB = 2R\sin\theta$ where $\theta=\angle APB=\pi/4$. So $8 = 2R\sin(45^\circ)=2R\cdot\frac{\sqrt2}{2}=R\sqrt2$, giving $R=\dfrac{8}{\sqrt2}=4\sqrt2$, so $R^2=32$.
    3. The center of such a circle lies on the perpendicular bisector of $AB$, which is the horizontal line $y=0$ through $M=(5,0)$. If the center is at distance $k$ from $M$ along this line, then (half-chord)$^2$ + $k^2$ = $R^2$: $4^2+k^2=32\Rightarrow k^2=16\Rightarrow k=\pm4$. …

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12. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.If the tangents drawn from a point P to the ellipse $4x^2+9y^2-16x+54y+61=0$ are perpendicular, then the locus of P is (A) $x^2+y^2-4x+6y+4=0$ (B) $x^2+y^2-4x+6y=0$ (C) $x^2+y^2-6x+4y+9=0$ (D) $x^2+y^2-6x+4y=0$

    ›Reveal solutionSolution

    The locus of the point of perpendicular tangents to an ellipse is its director circle $x^2+y^2=a^2+b^2$ (shifted to the ellipse's actual center here). The answer is $x^2+y^2-4x+6y=0$.

    Concept and Intuition

    Any ellipse has a director circle: the set of points from which the two tangents drawn to the ellipse are mutually perpendicular. For the standard ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, this circle is $x^2+y^2=a^2+b^2$, concentric with the ellipse. The reasoning: the pair of tangents from $(x_1,y_1)$ is $SS_1=T^2$; for the two lines represented by this homogeneous-in-direction equation to be perpendicular, the sum of coefficients of $x^2$ and $y^2$ must vanish, and working that out for the ellipse yields exactly $x_1^2+y_1^2=a^2+b^2$.

    Step-by-Step Solution

    1. Given: $4x^2+9y^2-16x+54y+61=0$. Group and complete the square. $4(x^2-4x)+9(y^2+6y)=-61$ $4[(x-2)^2-4]+9[(y+3)^2-9]=-61$ $4(x-2)^2+9(y+3)^2 = -61+16+81=36$.
    2. Divide by 36: $\dfrac{(x-2)^2}{9}+\dfrac{(y+3)^2}{4}=1$. So the ellipse has center $(2,-3)$, $a^2=9$, $b^2=4$.
    3. The director circle (locus of perpendicular-tangent points) is centered at the ellipse's center with $r^2=a^2+b^2=9+4=13$: $(x-2)^2+(y+3)^2=13$. …

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