Q.Given the ellipse with equation $9x^2 + 25y^2 = 225$, find the eccentricity and foci.

Showing the 12 most recent of 41 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $e$ is the eccentricity and $LL'$ is the length of the latus rectum of an ellipse $9x^2 + 4y^2 - 36x - 8y + 4 = 0$ then $e^2 LL' =$ (A) $\dfrac{40}{81}$ (B) $9$ (C) $5$ (D) $\dfrac{40}{27}$

   ›Reveal solutionSolution

   Completing the square reveals a standard ellipse with vertical major axis; computing $e^2$ and the latus rectum length gives $e^2LL'=40/27$.

   Concept and Intuition

   Any general second-degree ellipse equation can be reduced to standard form by completing the square in $x$ and $y$ separately, after which the standard formulas for eccentricity and latus rectum apply directly.

   Step-by-Step Solution

   1. $9x^2+4y^2-36x-8y+4=0 \Rightarrow 9(x^2-4x)+4(y^2-2y)+4=0$.
   2. $9[(x-2)^2-4]+4[(y-1)^2-1]+4=0 \Rightarrow 9(x-2)^2+4(y-1)^2-36-4+4=0 \Rightarrow 9(x-2)^2+4(y-1)^2=36$.
   3. Divide by 36: $\dfrac{(x-2)^2}{4}+\dfrac{(y-1)^2}{9}=1$. Since $9>4$, major axis is along $y$: $a^2=9,b^2=4$ (with $a=3,b=2$).
   4. $e^2=1-\dfrac{b^2}{a^2}=1-\dfrac49=\dfrac59$.
   5. Latus rectum length $LL'=\dfrac{2b^2}{a}=\dfrac{2\cdot4}{3}=\dfrac83$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the area of a triangle formed by the coordinate axes and the tangent drawn to the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at a point $\theta$ is $8\sqrt{3}$, then sum of all such values of $\theta$ is (A) $2\pi$ (B) $\pi$ (C) $\dfrac{3\pi}{2}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Writing the tangent at parameter $\theta$, the triangle-area condition reduces to $|\sin2\theta|=\sqrt3/2$; its four roots in $[0,2\pi)$ sum to $2\pi$.

   Concept and Intuition

   For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the point $(a\cos\theta, b\sin\theta)$ has tangent $\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1$. This line meets the axes at $\left(\dfrac{a}{\cos\theta},0\right)$ and $\left(0,\dfrac{b}{\sin\theta}\right)$, and the area of the right triangle these intercepts form with the origin is $\dfrac12\cdot\dfrac{a}{|\cos\theta|}\cdot\dfrac{b}{|\sin\theta|}$.

   Step-by-Step Solution

   1. Here $a=4,\ b=3$. Area $=\dfrac{1}{2}\cdot\dfrac{4}{|\cos\theta|}\cdot\dfrac{3}{|\sin\theta|}=\dfrac{6}{|\sin\theta\cos\theta|}=\dfrac{12}{|\sin2\theta|}$.
   2. Set equal to $8\sqrt3$: $\dfrac{12}{|\sin2\theta|}=8\sqrt3 \Rightarrow |\sin2\theta|=\dfrac{12}{8\sqrt3}=\dfrac{3}{2\sqrt3}=\dfrac{\sqrt3}{2}$.
   3. So $\sin2\theta=\pm\dfrac{\sqrt3}{2}$. For $2\theta\in[0,2\pi)$: $2\theta=\dfrac{\pi}{3},\dfrac{2\pi}{3},\dfrac{4\pi}{3},\dfrac{5\pi}{3}$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If the major axis of an ellipse subtends an angle of $120°$ at one end of its minor axis and the length of its semi latus rectum is $\dfrac{4}{\sqrt{3}}$, then the sum of the lengths of its axes is (A) $12$ (B) $24$ (C) $8(\sqrt{3} + 1)$ (D) $4(\sqrt{3} + 1)$

   ›Reveal solutionSolution

   The subtended-angle condition gives $a^2=3b^2$; combined with the semi-latus-rectum value this yields $b=4,\,a=4\sqrt3$, and the total axis length $8(\sqrt3+1)$.

   Concept and Intuition

   The angle subtended by the major axis at an end of the minor axis is a classic ellipse relation connecting $a$ and $b$ through the cosine rule in the isosceles triangle formed by the two vertices and the minor-axis end. Combined with the semi-latus-rectum formula $\ell=\dfrac{b^2}{a}$, we get two equations in $a,b$.

   Step-by-Step Solution

   1. Let $B=(0,b)$, $A_1=(-a,0)$, $A_2=(a,0)$. Then $\vec{BA_1}=(-a,-b)$, $\vec{BA_2}=(a,-b)$.
   2. $\cos(\angle A_1BA_2)=\dfrac{\vec{BA_1}\cdot\vec{BA_2}}{|\vec{BA_1}||\vec{BA_2}|}=\dfrac{-a^2+b^2}{a^2+b^2}$.
   3. Given the angle is $120^\circ$: $\dfrac{b^2-a^2}{a^2+b^2}=-\dfrac12 \Rightarrow 2(b^2-a^2)=-(a^2+b^2) \Rightarrow 3b^2=a^2$, i.e. $a=b\sqrt3$.
   4. Semi-latus rectum: $\dfrac{b^2}{a}=\dfrac{4}{\sqrt3}$. Substitute $a=b\sqrt3$: $\dfrac{b^2}{b\sqrt3}=\dfrac{b}{\sqrt3}=\dfrac{4}{\sqrt3}\Rightarrow b=4$. …

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4. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The tangents drawn at the points $P_1$ and $P_2$ lying on the ellipse $\dfrac{x^2}{4} + y^2 = 1$ are parallel to the chord joining the points $(0,1)$ and $(2,0)$, then the distance between $P_1$ and $P_2$ is (A) $2\sqrt{2}$ (B) $\sqrt{5}$ (C) $2\sqrt{3}$ (D) $\sqrt{10}$

   ›Reveal solutionSolution

   Finding the two tangents parallel to the given chord and their points of contact gives $P_1P_2=\sqrt{10}$. Answer: (D).

   Concept and Intuition

   A chord's slope tells us the direction of any tangent line parallel to it. For an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, tangents of slope $m$ are $y=mx\pm\sqrt{a^2m^2+b^2}$ — there are exactly two, symmetric about the center, touching at diametrically opposite points.

   Step-by-Step Solution

   1. Ellipse: $a^2=4, b^2=1$. Chord through $(0,1)$ and $(2,0)$ has slope $m=\dfrac{0-1}{2-0}=-\dfrac12$.
   2. Tangents with slope $-\dfrac12$: $c=\pm\sqrt{a^2m^2+b^2}=\pm\sqrt{4\cdot\frac14+1}=\pm\sqrt2$.
   3. Point of contact of $y=mx+c$ on the ellipse is $\left(\dfrac{-a^2m}{c},\dfrac{b^2}{c}\right)$. For $c=\sqrt2$: $P_1=\left(\dfrac{-4\cdot(-\frac12)}{\sqrt2},\dfrac{1}{\sqrt2}\right)=\left(\dfrac{2}{\sqrt2},\dfrac{1}{\sqrt2}\right)=(\sqrt2,\tfrac{\sqrt2}{2})$. For $c=-\sqrt2$: $P_2=(-\sqrt2,-\tfrac{\sqrt2}{2})$ (by symmetry, diametrically opposite $P_1$).
   4. Distance: …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.An ellipse intersects the hyperbola $2x^2 - 2y^2 = 1$ orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then the equation of that ellipse is (A) $x^2 + 2y^2 = 4$ (B) $2x^2 + y^2 = 4$ (C) $2x^2 + y^2 = 2$ (D) $x^2 + 2y^2 = 2$

   ›Reveal solutionSolution

   Orthogonal intersection with a given eccentricity ratio forces the ellipse and hyperbola to be confocal; solving the two conditions together gives $x^2+2y^2=2$. Answer: (D).

   Concept and Intuition

   When an ellipse and hyperbola share the same foci (confocal conics) and have axes along the same coordinate axes, they always intersect orthogonally — this is a classical fact. So "intersect orthogonally" here is equivalent to "confocal": ellipse's $c^2=A^2-B^2$ must equal hyperbola's $c^2=a^2+b^2$.

   Step-by-Step Solution

   1. Hyperbola: $2x^2-2y^2=1 \Rightarrow \dfrac{x^2}{1/2}-\dfrac{y^2}{1/2}=1$, so $a^2=b^2=\dfrac12$.
   2. Hyperbola's eccentricity: $e_h^2=1+\dfrac{b^2}{a^2}=1+1=2 \Rightarrow e_h=\sqrt2$.
   3. Ellipse's eccentricity is the reciprocal: $e=\dfrac{1}{\sqrt2} \Rightarrow e^2=\dfrac12$.
   4. Confocal condition (orthogonality with axes aligned): $A^2-B^2=a^2+b^2=\dfrac12+\dfrac12=1$. …

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6. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the distance between the foci of an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ is 6 and the distance between its directrices is 10, then the equation of one of the tangents of the ellipse drawn parallel to the line $y=\sqrt{2}x+5$ is (A) $y=\sqrt{2}x+\sqrt{66}$ (B) $y=\sqrt{2}x+12$ (C) $y=\sqrt{2}x+\sqrt{44}$ (D) $y=\sqrt{2}x+6$

   ›Reveal solutionSolution

   Using the focus/directrix data to pin down $a^2,b^2$ for the ellipse, then applying the tangent condition $c^2=a^2m^2+b^2$ for slope $\sqrt2$ gives $c=\pm6$; the answer is $y=\sqrt2x+6$.

   Concept and Intuition

   For an ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, the distance between the foci is $2ae$ and the distance between the directrices is $2a/e$. These two given lengths let us solve for $a^2$ and $e^2$ without ever separately knowing $a$ or $e$ — their product gives $a^2$ and their ratio gives $e^2$, which is a useful shortcut. Once $a^2,b^2$ are known, any line of a given slope that touches the ellipse exactly once (a tangent) must satisfy a fixed algebraic condition relating its intercept to $a,b,m$.

   Step-by-Step Solution

   1. Distance between foci $=2ae=6\Rightarrow ae=3$.
   2. Distance between directrices $=\dfrac{2a}{e}=10\Rightarrow \dfrac{a}{e}=5$.
   3. Multiply the two: $(ae)\left(\dfrac{a}{e}\right)=a^2=3\times5=15$.
   4. Divide: $\dfrac{ae}{a/e}=e^2=\dfrac{3}{5}$.
   5. Then $b^2=a^2(1-e^2)=15\left(1-\dfrac35\right)=15\cdot\dfrac25=6$. …

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7. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the ends of the major axis $A'$ and $A$ of the ellipse $\frac{(x-2)^2}{a^2} + \frac{(y-3)^2}{b^2} = 1$ are respectively at distances of $9$ and $3$ units from a directrix $L$, then the foci of the ellipse are (A) $\left(2 \pm \frac{3}{4}, 3\right)$ (B) $\left(2 \pm \frac{3}{2}, 3\right)$ (C) $\left(\pm \frac{3}{2}, 3\right)$ (D) $\left(\pm \frac{3}{4}, 3\right)$

   ›Reveal solutionSolution

   This tests the directrix-distance property of an ellipse's vertices. The two given distances give two equations in $a$ and $a/e$, and the foci follow from $e$ and the centre.

   Concept and Intuition

   For an ellipse with semi-major axis $a$ and eccentricity $e$, the distance from a vertex to the near directrix is $\frac{a}{e}-a$, and the distance from the opposite vertex (farther from that directrix) is $\frac{a}{e}+a$. Knowing both distances lets us solve for $a$ and $e$ independently of the actual coordinate values.

   Step-by-Step Solution

   1. Centre of the ellipse is $(2,3)$; major axis is horizontal, vertices $A=(2+a,3)$, $A'=(2-a,3)$.
   2. The nearer vertex is at distance $\frac{a}{e}-a$ from the directrix, the farther one at $\frac{a}{e}+a$. Given values are $3$ and $9$.
   3. $\frac{a}{e}-a=3$ and $\frac{a}{e}+a=9$.
   4. Adding: $\frac{2a}{e}=12\Rightarrow \frac{a}{e}=6$. …

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8. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.X - axis is the major axis and origin is the centre of an ellipse. If the distance between its directrices is $\dfrac{18}{\sqrt{5}}$ and the ratio between the distances from the centre of this ellipse to its focus and its corresponding directrices is 5 : 9, then the length of its latus rectum is (A) $\dfrac{8}{5}$ (B) $\dfrac{9}{5}$ (C) $\dfrac{8}{3}$ (D) $\dfrac{16}{3}$

   ›Reveal solutionSolution

   This tests the relations linking eccentricity, the focus-directrix ratio, and the directrix separation of an ellipse; the answer is $8/3$.

   Concept and Intuition

   For an ellipse with centre at the origin and major axis along the x-axis, the focus is at distance $ae$ from the centre and each directrix is at distance $a/e$ from the centre. The ratio of these two distances is therefore always $e^2$, regardless of $a$. The two directrices are symmetric about the centre, so the distance between them is $2a/e$.

   Step-by-Step Solution

   1. Ratio of (distance of focus from centre) to (distance of directrix from centre) $= \dfrac{ae}{a/e} = e^2 = \dfrac{5}{9}$.
   2. Distance between directrices $= \dfrac{2a}{e} = \dfrac{18}{\sqrt5}$, so $\dfrac{a}{e} = \dfrac{9}{\sqrt5}$.
   3. From $e^2=5/9$, $e=\dfrac{\sqrt5}{3}$. Substituting, $a = \dfrac{9}{\sqrt5}\cdot\dfrac{\sqrt5}{3} = 3$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The square of the slope of a common tangent drawn to the circle $4x^2 + 4y^2 = 25$ and the ellipse $4x^2 + 9y^2 = 36$ is (A) $1$ (B) $\dfrac{9}{11}$ (C) $\dfrac{2}{3}$ (D) $2$

   ›Reveal solutionSolution

   Equate the ellipse's tangent-line condition with the circle's tangency condition to get $m^2=9/11$.

   Concept and Intuition

   A line of slope $m$ touches the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ exactly when its intercept is $c=\pm\sqrt{a^2m^2+b^2}$. That same line is tangent to a circle centred at the origin exactly when its perpendicular distance from the origin equals the radius. A common tangent to both curves must satisfy both conditions simultaneously, which pins down $m$.

   Step-by-Step Solution

   1. Circle: $4x^2+4y^2=25\Rightarrow x^2+y^2=\dfrac{25}{4}$, so $r=\dfrac52$.
   2. Ellipse: $4x^2+9y^2=36\Rightarrow \dfrac{x^2}{9}+\dfrac{y^2}{4}=1$, so $a^2=9,b^2=4$.
   3. Tangent to ellipse of slope $m$: $y=mx+c$ with $c^2=a^2m^2+b^2=9m^2+4$.
   4. For this line to be tangent to the circle: $\dfrac{|c|}{\sqrt{1+m^2}}=r\Rightarrow c^2=r^2(1+m^2)$. …

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10. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If $ax^2 + 2hxy - 2ay^2 + 3x + 15y - 9 = 0$ represents a pair of lines intersecting at $(1,1)$, then $ah =$ (A) 14 (B) -15 (C) -7 (D) 9

    ›Reveal solutionSolution

    The point of intersection of a pair of lines makes both partial-derivative-style linear conditions vanish; applying them at $(1,1)$ solves for $a$ and $h$, giving $ah=-7$.

    Concept and Intuition

    For the general second-degree equation representing a pair of straight lines, $S \equiv Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$, the point of intersection $(x_0,y_0)$ of the two lines is exactly where $\dfrac{\partial S}{\partial x}=0$ and $\dfrac{\partial S}{\partial y}=0$ simultaneously, i.e. $Ax_0+Hy_0+G=0$ and $Hx_0+By_0+F=0$. This gives two linear equations in the unknown coefficients directly from the known intersection point.

    Step-by-Step Solution

    1. Match $ax^2+2hxy-2ay^2+3x+15y-9=0$ to the standard form: $A=a,\ H=h,\ B=-2a,\ 2G=3\Rightarrow G=\tfrac32,\ 2F=15\Rightarrow F=\tfrac{15}2,\ C=-9$.
    2. At $(x_0,y_0)=(1,1)$: $A(1)+H(1)+G=0\Rightarrow a+h+\tfrac32=0\Rightarrow a+h=-\tfrac32$.
    3. Also $H(1)+B(1)+F=0\Rightarrow h-2a+\tfrac{15}2=0\Rightarrow h-2a=-\tfrac{15}2$. …

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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.The angle between the tangents drawn from a point $(-3, 2)$ to the ellipse $4x^2 + 9y^2 - 36 = 0$ is (A) $45°$ (B) $\mathrm{Tan}^{-1}\left(\frac{2}{3}\right)$ (C) $\mathrm{Tan}^{-1}\left(\frac{3}{2}\right)$ (D) $90°$

    ›Reveal solutionSolution

    The combined pair-of-tangents equation from $(-3,2)$ to the ellipse factors neatly into $x=-3$ and $y=2$, two perpendicular lines. Answer: $90°$.

    Concept and Intuition

    From an external point $(x_1,y_1)$, the pair of tangent lines to a conic $S=0$ is given by $SS_1=T^2$, where $S_1$ is $S$ evaluated at the point and $T$ is the "polar" expression. This combined second-degree equation, when factored, gives the actual two tangent lines directly — from which the angle between them can be read off immediately (rather than computing it via a general angle-between-lines formula).

    Step-by-Step Solution

    1. Ellipse: $4x^2+9y^2-36=0 \Rightarrow \dfrac{x^2}{9}+\dfrac{y^2}{4}-1=0$, so $a^2=9,\ b^2=4$.
    2. $S_1 = \dfrac{(-3)^2}{9}+\dfrac{2^2}{4}-1 = 1+1-1=1$.
    3. $T = \dfrac{x(-3)}{9}+\dfrac{y(2)}{4}-1 = -\dfrac{x}{3}+\dfrac{y}{2}-1$.
    4. Pair of tangents: $S\cdot S_1 = T^2 \Rightarrow \dfrac{x^2}{9}+\dfrac{y^2}{4}-1 = \left(-\dfrac{x}{3}+\dfrac{y}{2}-1\right)^2$.
    5. Expanding the right side: $\dfrac{x^2}{9}+\dfrac{y^2}{4}+1-\dfrac{xy}{3}+\dfrac{2x}{3}-y$.
    6. Cancelling the common $\dfrac{x^2}{9}+\dfrac{y^2}{4}$ terms: $-1 = 1-\dfrac{xy}{3}+\dfrac{2x}{3}-y$, which rearranges to $xy-2x+3y-6=0$. …

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12. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
    Q.If a tangent having slope $\dfrac{1}{3}$ to the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$ (a > b) is a normal to the circle $(x+1)^2 + (y+1)^2 = 1$, then $a^2$ lies in the interval (A) $\left(\dfrac{\sqrt{2}}{\sqrt{5}}, 2\right)$ (B) $\left(\dfrac{2}{5}, 4\right)$ (C) $\left(1, \dfrac{10}{9}\right)$ (D) $(3, 5)$

    ›Reveal solutionSolution

    This tests the fact that a line is normal to a circle iff it passes through the circle's centre, combined with the tangent condition for an ellipse. Answer: $a^2\in\left(\dfrac25,4\right)$.

    Concept and Intuition

    Every diameter (line through the centre) of a circle is a normal to it at the two points it meets the circle, because the radius to any point of the circle is perpendicular to the tangent there, and a line through the centre is along that radius direction. So "this line is a normal to the circle" is just the geometric statement: the line passes through $(-1,-1)$.

    Separately, a line $y=mx+c$ touches the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ iff $c^2=a^2m^2+b^2$ — this is the standard tangency condition (discriminant of the intersection quadratic $=0$).

    Step-by-Step Solution

    1. The given tangent has slope $m=\dfrac13$ and must pass through the circle's centre $(-1,-1)$ (since it is a normal to the circle).
    2. Line through $(-1,-1)$ with slope $\dfrac13$: $y-(-1)=\dfrac13(x-(-1)) \Rightarrow y=\dfrac13x-\dfrac23$. So $c=-\dfrac23$.
    3. Tangency to the ellipse requires $c^2=a^2m^2+b^2$: $\left(\dfrac23\right)^2=a^2\left(\dfrac13\right)^2+b^2 \Rightarrow \dfrac49=\dfrac{a^2}{9}+b^2$.
    4. So $b^2=\dfrac{4-a^2}{9}$. …

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