Q.The base of an equilateral triangle with side $2a$ lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Showing the 12 most recent of 22 on this concept.

1. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.Let $O$ be the origin and $P$ be a point on a line such that $OP$ is perpendicular to that line. If $OP$ makes an obtuse angle $\alpha$ with the $x$-axis, $OP = 5$ and $\sin\alpha = \dfrac{3}{5}$ , then the equation of the line is (A) $3x - 4y - 25 = 0$ (B) $4x + 3y + 25 = 0$ (C) $3x - 4y + 25 = 0$ (D) $4x - 3y - 25 = 0$ (E) $4x - 3y + 25 = 0$

   ›Reveal solutionSolution

   The line is the perpendicular to $OP$ at its foot; use the normal form $x\cos\alpha+y\sin\alpha=p$ with $p=OP=5$.

   Since $OP$ is perpendicular to the line, $OP$ is the normal from the origin and $p=OP=5$. The angle is obtuse with $\sin\alpha=\tfrac35$, so $\cos\alpha=-\sqrt{1-\tfrac{9}{25}}=-\tfrac45$.

   The normal form of the line is $x\cos\alpha+y\sin\alpha=p$:

   $-\tfrac45 x+\tfrac35 y=5.$

   Multiplying by $5$: $-4x+3y=25$, i.e. $4x-3y+25=0$.

   ✓Final answer

   The correct option is (E).

2. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If the foci of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ coincide with the foci of the ellipse $\dfrac{x^2}{49} + \dfrac{y^2}{36} = 1$ , then the value of $a^2 + b^2$ is equal to (A) 25 (B) 36 (C) 85 (D) 169 (E) 13

   ›Reveal solutionSolution

   The shared focal distance $c$ satisfies $c^2=13$; for the hyperbola $a^2+b^2=c^2$.

   For the ellipse $\dfrac{x^2}{49}+\dfrac{y^2}{36}=1$, $c^2=49-36=13$.

   The hyperbola's foci coincide, so its focal parameter is the same: for $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $c^2=a^2+b^2=13$.

   Therefore $a^2+b^2=13$.

   ✓Final answer

   The correct option is (E).

3. KEAM 2026Set eng-2026-04174 marksMCQ
   Q.If the distance between the foci of an ellipse is 4 and its eccentricity is $\dfrac{1}{2}$ , then the length of its latus rectum is (A) 4 (B) 8 (C) 6 (D) 12 (E) 10

   ›Reveal solutionSolution

   From $2c=4$ and $e=\tfrac12$ get $a,b^2$, then $\dfrac{2b^2}{a}$.

   Distance between foci $2c=4\Rightarrow c=2$. Eccentricity $e=\dfrac{c}{a}=\dfrac12\Rightarrow a=4$.

   $b^2=a^2-c^2=16-4=12$.

   Length of latus rectum $=\dfrac{2b^2}{a}=\dfrac{2\cdot12}{4}=6$.

   ✓Final answer

   The correct option is (C).

4. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.A point $C$ lies on the perpendicular bisector of the straight-line segment joining the points $A(-3,-6)$ and $B(13,-6)$. If the point $C$ lies in the first quadrant and the distance between the point $C$ and the midpoint of $AB$ is 8 units, then the coordinates of $C$ are (A) $(5,1)$ (B) $(5,2)$ (C) $(5,3)$ (D) $(5,4)$ (E) $(5,5)$

   ›Reveal solutionSolution

   The perpendicular bisector is $x=5$; the point 8 units from $(5,-6)$ in the first quadrant is $(5,2)$.

   $A(-3,-6)$ and $B(13,-6)$ share $y=-6$, so their midpoint is $M=(5,-6)$ and the perpendicular bisector is the vertical line $x=5$.

   Points on it at distance $8$ from $M$: $(5,-6\pm8)=(5,2)$ or $(5,-14)$. The first-quadrant point is $C=(5,2)$.

   ✓Final answer

   The correct option is (B).

5. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.If the eccentricity and the length of latus rectum of an ellipse are, respectively, $\dfrac{1}{5}$ and $\dfrac{48}{5}$, then the length of the major axis of the ellipse is (A) $5$ (B) $6$ (C) $8$ (D) $10$ (E) $12$

   ›Reveal solutionSolution

   Combine LR $=\tfrac{2b^2}{a}$ with $b^2=a^2(1-e^2)$ to get $a=5$, so major axis $2a=10$.

   With $e=\tfrac15$, $b^{2}=a^{2}(1-e^{2})=a^{2}\big(1-\tfrac{1}{25}\big)=\tfrac{24}{25}a^{2}$.

   Latus rectum $=\dfrac{2b^{2}}{a}=\dfrac{2\cdot\tfrac{24}{25}a^{2}}{a}=\dfrac{48a}{25}$. Setting this equal to $\tfrac{48}{5}$:

   $\frac{48a}{25}=\frac{48}{5}\;\Rightarrow\;a=5.$

   Major axis $=2a=10$.

   ✓Final answer

   The correct option is (D).

6. KEAM 2026Set eng-2026-04184 marksMCQ
   Q.The latus rectum of the hyperbola $\dfrac{(3x-7)^{2}}{9}-\dfrac{(4y+3)^{2}}{8}=1$ is (A) $\dfrac{1}{2}$ (B) $\dfrac{16}{3}$ (C) $4$ (D) $\dfrac{4}{3}$ (E) $1$

   ›Reveal solutionSolution

   Normalizing to $(x-\tfrac73)^2/1-(y+\tfrac34)^2/\tfrac12=1$ gives $a^2=1$, $b^2=\tfrac12$, so LR $=1$.

   Write $3x-7=3(x-\tfrac73)$ and $4y+3=4(y+\tfrac34)$:

   $\frac{9(x-\tfrac73)^2}{9}-\frac{16(y+\tfrac34)^2}{8}=\frac{(x-\tfrac73)^2}{1}-\frac{(y+\tfrac34)^2}{1/2}=1.$

   So $a^{2}=1$ and $b^{2}=\tfrac12$.

   Latus rectum $=\dfrac{2b^{2}}{a}=\dfrac{2\cdot\tfrac12}{1}=1$.

   ✓Final answer

   The correct option is (E).

7. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If the length of the major axis of an ellipse is thrice the length of the minor axis, then its eccentricity is equal to (A) $\frac{2\sqrt{2}}{3}$ (B) $\frac{\sqrt{2}}{\sqrt{3}}$ (C) $\frac{1}{2}$ (D) $\frac{1}{\sqrt{2}}$ (E) $\frac{1}{2\sqrt{2}}$

   ›Reveal solutionSolution

   Major axis $=3\times$minor axis means $a=3b$, giving $e=\frac{2\sqrt2}{3}$.

   Length of major axis $=2a$, minor axis $=2b$. Given $2a=3(2b)$, so $a=3b$. Then

   $e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{b^2}{9b^2}}=\sqrt{1-\frac19}=\sqrt{\frac89}=\frac{2\sqrt2}{3}.$

   ✓Final answer

   The correct option is (A).

8. KEAM 2026Set eng-2026-04214 marksMCQ
   Q.If $O$ is the origin and $C$ is the midpoint of $A(-2,1)$ and $B(4,-3)$, then $\overrightarrow{OC}$ is (A) $\hat{i}+\hat{j}$ (B) $-\hat{i}+-\hat{j}$ (C) $\frac{1}{2}\hat{i}+\frac{1}{2}\hat{j}$ (D) $\hat{i}-\hat{j}$ (E) $-\hat{i}+\hat{j}$

   ›Reveal solutionSolution

   The midpoint of $A(-2,1)$ and $B(4,-3)$ is $(1,-1)$, giving $\overrightarrow{OC}=\hat i-\hat j$.

   Midpoint:

   $C=\left(\frac{-2+4}{2},\frac{1+(-3)}{2}\right)=(1,-1).$

   With $O$ the origin, $\overrightarrow{OC}=\hat i-\hat j$.

   ✓Final answer

   The correct option is (D).

9. KEAM 2026Set eng-2026-04224 marksMCQ
   Q.The $x$-intercept and $y$-intercept of a line are three times and four times of the $x$-intercept and $y$-intercept of the line $3x + 2y = 6$, respectively. Then the equation of the line is (A) $2x - y = 12$ (B) $2x + y = 12$ (C) $2x - y = -12$ (D) $x + 2y = 12$ (E) $2x + 2y = 12$

   ›Reveal solutionSolution

   Get the base line's intercepts, scale them, then form the intercept-form equation.

   For $3x+2y=6$: x-intercept $=2$ (set $y=0$), y-intercept $=3$ (set $x=0$).

   New x-intercept $=3\times2=6$; new y-intercept $=4\times3=12$.

   Intercept form: $\frac{x}{6}+\frac{y}{12}=1$. Multiply by $12$: $2x+y=12$.

   ✓Final answer

   The correct option is (B).

10. KEAM 2026Set eng-2026-04224 marksMCQ
    Q.If the foci and vertices of an ellipse are respectively $(\pm 2, 0)$ and $(\pm 3, 0)$ then its eccentricity is (A) $\frac{2}{3}$ (B) $\frac{\sqrt{5}}{3}$ (C) $\frac{\sqrt{2}}{3}$ (D) $\frac{1}{2}$ (E) $\frac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    Read $a$ from the vertices and $c$ from the foci; eccentricity is $c/a$.

    For the ellipse, $a=3$ (vertices $(\pm3,0)$) and $c=2$ (foci $(\pm2,0)$).

    Eccentricity $e=\frac{c}{a}=\frac{2}{3}$.

    ✓Final answer

    The correct option is (A).

11. KEAM 2026Set eng-2026-04224 marksMCQ
    Q.The equation of a hyperbola is $9x^{2} - 16y^{2} = 144$. If $A$ and $S$ are, respectively, the focus and the vertex of one section of the hyperbola, then the length of $AS$ is (A) $\frac{5}{2}$ (B) $\frac{3}{2}$ (C) $\frac{1}{2}$ (D) 2 (E) 1

    ›Reveal solutionSolution

    Put the hyperbola in standard form, find $a$ and $c$, then $AS=c-a$.

    $9x^2-16y^2=144\Rightarrow\frac{x^2}{16}-\frac{y^2}{9}=1$, so $a^2=16$, $b^2=9$.

    $c^2=a^2+b^2=25\Rightarrow c=5$; vertex at $(4,0)$, focus at $(5,0)$.

    Length $AS=5-4=1$.

    ✓Final answer

    The correct option is (E).

12. KEAM 2025Set eng-2025-04234 marksMCQ
    Q.The length of major axis and minor axis of an ellipse are, respectively, $m$ and $n$. If $m^2 - n^2 = 45$ and the eccentricity of the ellipse is $\dfrac{\sqrt{5}}{3}$, then the length of the major axis is (A) $13$ (B) $6$ (C) $12$ (D) $18$ (E) $9$

    ›Reveal solutionSolution

    The major axis length is $9$.

    Concept and Intuition

    With major axis $m=2a$ and minor axis $n=2b$, $m^2-n^2=4(a^2-b^2)=4a^2e^2$. Use the given eccentricity to solve for $a$.

    Step-by-Step Solution

    1. $m^2-n^2=4a^2-4b^2=4(a^2-b^2)=4a^2e^2$.
    2. $e^2=5/9$, so $4a^2\cdot\tfrac{5}{9}=45\Rightarrow \tfrac{20a^2}{9}=45\Rightarrow a^2=20.25$, $a=4.5$.
    3. Major axis $m=2a=9$.

    Common Mistakes

    • Confusing $m,n$ (axis lengths) with $a,b$ (semi-axes).
    • Using $b^2=a^2e^2$ instead of $a^2-b^2=a^2e^2$.

    ✓Final answer

    The correct option is (E) — $9$.

    ANSWER: E