Q.Show that the path of a moving point such that its distances from two lines $3x - 2y = 5$ and $3x + 2y = 5$ are equal is a straight line.

Showing the 12 most recent of 30 on this concept.

1. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $P = \left(\frac{15}{2}(cosec\,\theta + sin\,\theta), 8(cosec\,\theta - sin\,\theta)\right)$, where $\theta$ is a variable parameter. Then the locus of $P$ is (A) $\frac{x^{2}}{15} - \frac{y^{2}}{16} = 1$ (B) $\frac{x^{2}}{256} - \frac{y^{2}}{225} = 1$ (C) $\frac{x^{2}}{225} + \frac{y^{2}}{256} = 1$ (D) $\frac{x^{2}}{225} - \frac{y^{2}}{256} = 1$ (E) $\frac{x^{2}}{16} + \frac{y^{2}}{30} = 1$

   ›Reveal solutionSolution

   Form $(\csc\theta+\sin\theta)^2-(\csc\theta-\sin\theta)^2=4\csc\theta\sin\theta=4$ to eliminate $\theta$.

   From the coordinates, $\tfrac{2x}{15}=\csc\theta+\sin\theta$ and $\tfrac{y}{8}=\csc\theta-\sin\theta$. …

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2. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.The vertex of a parabola is at $(2,-5)$ and the focus is at $(5,-5)$. The equation of the parabola is (A) $y^{2} + 10y - 10x + 49 = 0$ (B) $y^{2} + 10y - 12x + 49 = 0$ (C) $y^{2} + 10y - 12x + 46 = 0$ (D) $y^{2} + 8y - 12x + 49 = 0$ (E) $y^{2} + 10y - 18x + 48 = 0$

   ›Reveal solutionSolution

   Focus lies to the right of the vertex, so the axis is horizontal with $a=3$; expand $(y+5)^2=12(x-2)$.

   Vertex $(2,-5)$, focus $(5,-5)$: same $y$, so the parabola opens rightward with $a=5-2=3$.

   $(y+5)^2=4a(x-2)=12(x-2).$ …

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3. KEAM 2026Set eng-2026-04194 marksMCQ
   Q.Let $R(-2, -2)$ be a point and let $\frac{(x - 3)^{2}}{25} + \frac{(y + 2)^{2}}{16} = 1$ be an ellipse. If $S$ and $T$ are the foci of the ellipse, then $RS + RT$ is equal to (A) 128 (B) 61 (C) 12 (D) 10 (E) 124

   ›Reveal solutionSolution

   Check that $R$ is on the ellipse; then the focal-distance sum is the constant $2a$.

   For $\dfrac{(x-3)^2}{25}+\dfrac{(y+2)^2}{16}=1$: at $R(-2,-2)$, $\dfrac{(-5)^2}{25}+\dfrac{0}{16}=1$, so $R$ is on the ellipse. …

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4. KEAM 2026Set eng-2026-04204 marksMCQ
   Q.Let the eccentricity of an ellipse be $\frac{1}{2}$. If $S(3,2)$ is a focus and $x-9=0$ is the corresponding directrix of the ellipse, then the equation of the ellipse is (A) $3x^2+4y^2-6x-16y-29=0$ (B) $3x^2+4y^2-8x-16y-29=0$ (C) $3x^2+4y^2-6x-12y-29=0$ (D) $3x^2+4y^2-6x-16y+29=0$ (E) $13x^2+5y^2-6x-16y-29=0$

   ›Reveal solutionSolution

   Using the focus $(3,2)$, directrix $x=9$, and $e=\frac12$: $\sqrt{(x-3)^2+(y-2)^2}=\frac12|x-9|$. Squaring, $4[(x-3)^2+(y-2)^2]=(x-9)^2$, which simplifies to $3x^2+4y^2-6x-16y-29=0$.

   By the focus–directrix definition, distance to focus $=e\times$ distance to directrix:

   $\sqrt{(x-3)^2+(y-2)^2}=\frac12\,|x-9|.$

   Square both sides:

   $(x-3)^2+(y-2)^2=\frac14(x-9)^2\;\Rightarrow\;4\left[(x-3)^2+(y-2)^2\right]=(x-9)^2.$

   Expand the left side: …

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5. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The foci of an ellipse are at $(-3,0)$ and $(3,0)$. If the eccentricity of the ellipse is $\frac{1}{2}$, then the equation of the ellipse is (A) $\frac{x^2}{25} + \frac{y^2}{16} = 1$ (B) $\frac{x^2}{16} + \frac{y^2}{7} = 1$ (C) $\frac{x^2}{16} + \frac{y^2}{25} = 1$ (D) $\frac{x^2}{36} + \frac{y^2}{16} = 1$ (E) $\frac{x^2}{36} + \frac{y^2}{27} = 1$

   ›Reveal solutionSolution

   From foci get $c=3$; eccentricity gives $a=6$; then $b^2=a^2-c^2=27$, giving $\tfrac{x^2}{36}+\tfrac{y^2}{27}=1$.

   Foci at $(\pm 3, 0)$ lie on the $x$-axis, so the major axis is horizontal with $c = 3$.

   Eccentricity $e = \dfrac{c}{a} = \dfrac12 \Rightarrow a = 2c = 6$, so $a^2 = 36$. …

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6. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The coordinates of the focus and the vertex of a parabola, respectively, are $(-1,4)$ and $(3,4)$. Then the equation of the parabola is (A) $(x-3)^2 = 16(y-4)$ (B) $(x-3)^2 = -16(y-4)$ (C) $(y-4)^2 = -8(x-3)$ (D) $(y-4)^2 = -16(x-3)$ (E) $(y-4)^2 = 1(x-3)$

   ›Reveal solutionSolution

   Vertex and focus share $y=4$ (horizontal axis); the focus is $4$ units left of the vertex, so the parabola opens left with $4a=16$.

   Focus $(-1,4)$ and vertex $(3,4)$ have the same $y$-coordinate, so the axis is horizontal ($y=4$). …

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7. KEAM 2025Set eng-2025-04254 marksMCQ
   Q.The line segment joining the points $(-3,1)$ and $(1,1)$ is transverse axis of a hyperbola. If the length of the conjugate axis is 4, then the equation of the hyperbola is (A) $(x+2)^2 - (y-1)^2 = 4$ (B) $(x+1)^2 - (y-1)^2 = 16$ (C) $(x+1)^2 - (y-1)^2 = 4$ (D) $(x+2)^2 - (y-1)^2 = 16$ (E) $(x-1)^2 - (y+1)^2 = 4$

   ›Reveal solutionSolution

   The transverse-axis endpoints give centre $(-1,1)$ and $a=2$; the conjugate axis gives $b=2$; the hyperbola is $(x+1)^2-(y-1)^2=4$.

   Transverse axis joins $(-3,1)$ and $(1,1)$: horizontal, midpoint (centre) $(-1,1)$, length $4$, so $a = 2$, $a^2 = 4$.

   Conjugate axis length $= 4 \Rightarrow 2b = 4 \Rightarrow b = 2$, $b^2 = 4$. …

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8. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.Let $P$ be any point on the ellipse $4(x + 2)^{2} + 9(y - 4)^{2} = 144$. If $F_{1}$ and $F_{2}$ are the Foci of the ellipse, then $F_{1}P + F_{2}P =$ (A) $8$ (B) $12$ (C) $16$ (D) $6$ (E) $10$

   ›Reveal solutionSolution

   Dividing by $144$ gives semi-major axis $a=6$; the sum of focal distances of any point on an ellipse equals $2a=12$.

   Rewrite $4(x+2)^2+9(y-4)^2=144$ by dividing by $144$:

   $\dfrac{(x+2)^2}{36}+\dfrac{(y-4)^2}{16}=1.$ …

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9. KEAM 2025Set eng-2025-04264 marksMCQ
   Q.The eccentricity of the hyperbola $\dfrac{(x-1)^{2}}{25} - \dfrac{(y+2)^{2}}{11} = 1$ is (A) $\dfrac{5}{3}$ (B) $\dfrac{25}{11}$ (C) $\dfrac{6}{5}$ (D) $\dfrac{7}{5}$ (E) $\dfrac{5}{11}$

   ›Reveal solutionSolution

   For a hyperbola $\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=1$ the eccentricity is $e=\sqrt{1+b^2/a^2}$; with $a^2=25,\,b^2=11$ this is $\dfrac{6}{5}$.

   Here $a^2=25$ and $b^2=11$. The eccentricity formula gives …

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10. KEAM 2025Set eng-2025-04274 marksMCQ
    Q.Length of the Latus rectum of the ellipse $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$ is (A) $\dfrac{3}{2}$ (B) $8$ (C) $\dfrac{9}{2}$ (D) $2$ (E) $\dfrac{25}{2}$

    ›Reveal solutionSolution

    With $a^2=16$, $b^2=9$, $a=4$, the latus rectum is $\dfrac{2b^2}{a}=\dfrac{9}{2}$.

    For $\dfrac{x^2}{9}+\dfrac{y^2}{16}=1$, the larger denominator $16$ is under $y^2$, so the major axis is along $y$ with $a^2=16$ (semi-major $a=4$) and $b^2=9$. …

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11. KEAM 2025Set eng-2025-04274 marksMCQ
    Q.The line $x-y+4=0$ touches the ellipse $x^2+3y^2=12$ at (A) $(1,3)$ (B) $(3,1)$ (C) $(0,2)$ (D) $(0,-2)$ (E) $(-3,1)$

    ›Reveal solutionSolution

    The line $y=x+4$ meets the ellipse in a repeated root $x=-3$, giving the point of tangency $(-3,1)$.

    From $x-y+4=0$ we get $y=x+4$. Substitute into $x^2+3y^2=12$:

    $x^2+3(x+4)^2=12\Rightarrow x^2+3x^2+24x+48=12.$ …

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12. KEAM 2025Set eng-2025-04284 marksMCQ
    Q.If the length of the latus rectum of an ellipse is one-fourth of the major axis, then the eccentricity of the ellipse is (A) $\dfrac{\sqrt{3}}{2}$ (B) $\dfrac{\sqrt{3}}{4}$ (C) $\dfrac{\sqrt{5}}{4}$ (D) $\dfrac{\sqrt{5}}{6}$ (E) $\dfrac{2}{3}$

    ›Reveal solutionSolution

    Set latus rectum $=\tfrac14\times$ major axis to find $b^2=\tfrac{a^2}{4}$, giving eccentricity $\tfrac{\sqrt3}{2}$.

    For an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ ($a>b$), the latus rectum length is $\dfrac{2b^2}{a}$ and the major axis is $2a$.

    Given condition: …

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