Q.Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola $\frac{y^2}{9} - \frac{x^2}{27} = 1$.

Showing the 12 most recent of 69 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.Let X-axis be the transverse axis and origin be the centre of a hyperbola. If its latus rectum subtends an angle of $150^\circ$ at its vertex, then its eccentricity is (A) $\sqrt{3} + 1$ (B) $2 + \sqrt{3}$ (C) $1 + \sqrt{2}$ (D) $2 + \sqrt{2}$

   ›Reveal solutionSolution

   Half the subtended angle gives $\tan(\theta/2)=e+1$ for a hyperbola's latus rectum viewed from its vertex; solving with $\theta=150^\circ$ gives $e=\sqrt3+1$.

   Concept and Intuition

   The latus rectum through a focus $(ae,0)$ has endpoints $(ae,\pm b^2/a)$. Viewed from the nearer vertex $(a,0)$, symmetry about the $X$-axis means the full subtended angle is twice the angle to one endpoint, and this angle simplifies neatly in terms of the eccentricity alone.

   Step-by-Step Solution

   1. Hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, vertex $(a,0)$, right focus $(ae,0)$, latus rectum endpoints $(ae,\pm b^2/a)$.
   2. Half-angle at vertex to one endpoint: $\tan(75^\circ)=\dfrac{b^2/a}{ae-a}=\dfrac{b^2}{a^2(e-1)}$.
   3. Using $b^2=a^2(e^2-1)$: $\tan75^\circ=\dfrac{a^2(e^2-1)}{a^2(e-1)}=\dfrac{(e-1)(e+1)}{e-1}=e+1$.
   4. $\tan75^\circ=\tan(45^\circ+30^\circ)=2+\sqrt3$ (standard value).
   5. So $e+1=2+\sqrt3\Rightarrow e=1+\sqrt3$.

   Common Mistakes

   • Using the full angle $150^\circ$ directly in the tangent formula instead of the half-angle $75^\circ$ (the vertex sees the two endpoints symmetric about the axis).
   • Misremembering $\tan75^\circ$ (it is $2+\sqrt3$, not $2-\sqrt3$, which is $\tan15^\circ$).

   ✓Final answer

   The correct option is (A) — $\sqrt3+1$.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.Let S, S' be the foci of a standard hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ and its eccentricity be $\sqrt{5}$. If 'b' is the radius of a circle concentric with the hyperbola, then the number of points of intersection of the circle and the hyperbola is (A) $4$ (B) $2$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   The circle of radius $b=2a$ concentric with the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ (with $e=\sqrt5$) intersects it at exactly $4$ real points.

   Concept and Intuition

   For a hyperbola, $b^2=a^2(e^2-1)$. Here $e=\sqrt5$ gives $b^2=4a^2$, i.e. the circle's radius $b=2a$ is bigger than the hyperbola's vertex distance $a$, so we expect the circle to genuinely cross both branches — the question is exactly how many points that produces.

   Step-by-Step Solution

   1. $b^2 = a^2(e^2-1) = a^2(5-1)=4a^2$, so radius of circle $=b=2a$, circle: $x^2+y^2=4a^2$.
   2. From the hyperbola: $x^2 = a^2\left(1+\dfrac{y^2}{b^2}\right) = a^2+\dfrac{a^2y^2}{4a^2}=a^2+\dfrac{y^2}{4}$.
   3. Substitute into the circle: $a^2+\dfrac{y^2}{4}+y^2=4a^2 \Rightarrow \dfrac{5y^2}{4}=3a^2 \Rightarrow y^2=\dfrac{12a^2}{5}$.
   4. This is positive, giving two real values $y=\pm\sqrt{\tfrac{12a^2}{5}}$.
   5. Then $x^2=a^2+\dfrac14\cdot\dfrac{12a^2}{5}=a^2+\dfrac{3a^2}{5}=\dfrac{8a^2}{5}$, also positive, giving $x=\pm\sqrt{\tfrac{8a^2}{5}}$.
   6. Since both the hyperbola equation and circle equation are even in $x$ and in $y$ separately, all four sign combinations $(\pm x,\pm y)$ satisfy both simultaneously — giving $4$ distinct real intersection points.

   Common Mistakes

   • Assuming a circle can meet a hyperbola in at most $2$ points (a hyperbola has two branches, each of which a circle can cross twice).
   • Sign errors when substituting $x^2$ from the hyperbola equation into the circle equation.

   ✓Final answer

   The correct option is (A) — $4$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.The equation of the conjugate hyperbola of the hyperbola $x^2 - 4y^2 - 2x - 8y - 19 = 0$ is (A) $x^2 - 4y^2 - 2x - 8y - 33 = 0$ (B) $x^2 - 4y^2 - 2x - 8y + 33 = 0$ (C) $x^2 - 4y^2 - 2x - 8y + 13 = 0$ (D) $x^2 - 4y^2 - 2x - 8y - 13 = 0$

   ›Reveal solutionSolution

   Completing the square gives the hyperbola $(x-1)^2-4(y+1)^2=16$; its conjugate flips the sign of the right side to $-16$, expanding to $x^2-4y^2-2x-8y+13=0$.

   Concept and Intuition

   For a hyperbola $\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=1$, the conjugate hyperbola is $\dfrac{Y^2}{b^2}-\dfrac{X^2}{a^2}=1$ — equivalently, keep the same quadratic/linear terms but negate the constant on the right (or, in the "$=k$" form, replace $k$ by $-k$), about the same centre.

   Step-by-Step Solution

   1. Group and complete the square: $x^2-2x-4(y^2+2y)-19=0 \Rightarrow (x-1)^2-1-4[(y+1)^2-1]-19=0$.
   2. $\Rightarrow (x-1)^2-1-4(y+1)^2+4-19=0 \Rightarrow (x-1)^2-4(y+1)^2-16=0 \Rightarrow (x-1)^2-4(y+1)^2=16$.
   3. This is $\dfrac{(x-1)^2}{16}-\dfrac{(y+1)^2}{4}=1$ — centre $(1,-1)$, $a^2=16,\,b^2=4$.
   4. The conjugate hyperbola replaces $16$ by $-16$: $(x-1)^2-4(y+1)^2=-16$, i.e. $(x-1)^2-4(y+1)^2+16=0$.
   5. Expand: $x^2-2x+1-4(y^2+2y+1)+16=0 \Rightarrow x^2-2x+1-4y^2-8y-4+16=0 \Rightarrow x^2-4y^2-2x-8y+13=0$.

   Common Mistakes

   • Forgetting to re-expand around the same centre (the linear terms $-2x,-8y$ stay identical for both original and conjugate — only the constant term changes).
   • Sign slip while completing the square for the $y$ term (factor of $-4$ must be distributed carefully).

   ✓Final answer

   The correct option is (C) — $x^2-4y^2-2x-8y+13=0$.

   ANSWER: C

4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.From a point $P(x_1,-1)$ $(x_1<0)$, two tangents are drawn to the hyperbola $\dfrac{x^2}{2}-\dfrac{y^2}{3}=1$. If the sum of the slopes of the tangents is 2, then $x_1=$ (A) $-1$ (B) $-2$ (C) $-\dfrac{1}{3}$ (D) $-\dfrac{2}{5}$

   ›Reveal solutionSolution

   Writing the tangent-line quadratic in slope $m$ for tangents from $P(x_1,-1)$ to the hyperbola, and using "sum of roots $=2$" as the given condition, yields $x_1=-2$.

   Concept and Intuition

   From an external point, exactly two tangent lines touch a conic; their slopes are the two roots of a quadratic obtained by substituting the tangency condition into the fact that the line passes through the point. Vieta's formulas then let us use "sum of slopes" or "product of slopes" as an equation in the point's coordinates — we never need to find the individual tangent lines.

   Step-by-Step Solution

   1. For $\dfrac{x^2}{2}-\dfrac{y^2}{3}=1$ ($a^2=2,b^2=3$), a line $y=mx+c$ is tangent iff $c^2=a^2m^2-b^2=2m^2-3$.
   2. The tangent passes through $P(x_1,-1)$: $-1=mx_1+c\Rightarrow c=-1-mx_1$.
   3. Substitute: $(-1-mx_1)^2=2m^2-3$.
   4. Expand: $1+2mx_1+m^2x_1^2=2m^2-3$.
   5. Rearrange: $m^2(x_1^2-2)+2x_1m+4=0$.
   6. This quadratic's two roots are the slopes $m_1,m_2$ of the two tangents. By Vieta, $m_1+m_2=\dfrac{-2x_1}{x_1^2-2}$.
   7. Given $m_1+m_2=2$: $\dfrac{-2x_1}{x_1^2-2}=2\Rightarrow -2x_1=2x_1^2-4\Rightarrow 2x_1^2+2x_1-4=0\Rightarrow x_1^2+x_1-2=0$.
   8. Factor: $(x_1+2)(x_1-1)=0\Rightarrow x_1=-2$ or $x_1=1$.
   9. Since $x_1<0$ is given, $x_1=-2$.

   Common Mistakes

   • Using the ellipse-style tangent condition ($+b^2$) instead of the hyperbola one ($-b^2$).
   • Forgetting to discard the root $x_1=1$ that violates $x_1<0$.

   ✓Final answer

   The correct option is (B) — $x_1=-2$.

   ANSWER: B

5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If the tangent and normal at any point on the hyperbola $x^2-y^2=a^2$ cut off intercepts $a_1$ and $a_2$ on X-axis, $b_1$ and $b_2$ on Y-axis respectively, then (A) $a_1 a_2 = b_1 b_2$ (B) $a_1 b_2 = a_2 b_1$ (C) $a_1 b_2 + a_2 b_1 = 0$ (D) $a_1 a_2 + b_1 b_2 = 0$

   ›Reveal solutionSolution

   Relates the axis-intercepts of the tangent and normal to a rectangular hyperbola; the identity is $a_1a_2+b_1b_2=0$.

   Concept and Intuition

   On a rectangular hyperbola $x^2-y^2=a^2$, the tangent and normal at a point have slopes that are negative reciprocals of each other's coordinate ratio in a special way, which forces a clean symmetric relation between their intercepts.

   Step-by-Step Solution

   1. Let the point be $(x_1,y_1)$ with $x_1^2-y_1^2=a^2$.
   2. Tangent: differentiating implicitly, $2x-2yy'=0\Rightarrow y'=x/y$, so tangent slope at the point is $x_1/y_1$. Tangent: $xx_1-yy_1=a^2$.
      • $x$-intercept $a_1$ (set $y=0$): $a_1=a^2/x_1$.
      • $y$-intercept $b_1$ (set $x=0$): $b_1=-a^2/y_1$.
   3. Normal: slope $=-y_1/x_1$ (negative reciprocal). Line: $y-y_1=-\dfrac{y_1}{x_1}(x-x_1)$.
      • $x$-intercept $a_2$ (set $y=0$): $-y_1=-\dfrac{y_1}{x_1}(x-x_1)\Rightarrow x-x_1=x_1\Rightarrow a_2=2x_1$.
      • $y$-intercept $b_2$ (set $x=0$): $y-y_1=\dfrac{y_1}{x_1}\cdot x_1=y_1\Rightarrow b_2=2y_1$.
   4. Compute $a_1a_2=\dfrac{a^2}{x_1}\cdot 2x_1=2a^2$ and $b_1b_2=-\dfrac{a^2}{y_1}\cdot 2y_1=-2a^2$.
   5. Hence $a_1a_2+b_1b_2=2a^2-2a^2=0$.

   Common Mistakes

   • Sign error on the $y$-intercept of the tangent (it is $-a^2/y_1$, not $+a^2/y_1$).
   • Assuming $a_1a_2=b_1b_2$ (option A) instead of checking the actual signs.

   ✓Final answer

   The correct option is (D) — $a_1a_2+b_1b_2=0$.

   ANSWER: D

6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Consider the Hyperbola $S \equiv \frac{x^2}{25} - \frac{y^2}{16} - 1 = 0$. Let $B, B'$ be the ends of the transverse axis of the conjugate hyperbola of $S = 0$. If $C$ is the circle with $B, B'$ as ends of diameter, then slope of a common tangent to $C$ and the given Hyperbola is (A) $\pm \frac{3\sqrt{2}}{4}$ (B) $\pm \frac{4\sqrt{2}}{3}$ (C) $\pm \frac{5\sqrt{3}}{4}$ (D) $\pm \frac{3\sqrt{3}}{2}$

   ›Reveal solutionSolution

   This tests tangent-line conditions to a hyperbola and a circle simultaneously. Identifying the conjugate hyperbola's transverse-axis circle and equating both tangent-constant formulas gives the common slope.

   Concept and Intuition

   The conjugate of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$, whose transverse axis lies along the y-axis with half-length $b$. A common tangent to two curves must have the same slope-form equation for both, so equating the two tangent-condition expressions in $m$ solves for the shared slope.

   Step-by-Step Solution

   1. For $S:\frac{x^2}{25}-\frac{y^2}{16}=1$, $a^2=25,b^2=16$.
   2. Conjugate hyperbola: $\frac{y^2}{16}-\frac{x^2}{25}=1$; its transverse axis is along the y-axis with half-length $\sqrt{16}=4$, so $B=(0,4),B'=(0,-4)$.
   3. Circle $C$ on diameter $BB'$: centre origin, radius $4$, i.e. $x^2+y^2=16$.
   4. Tangent of slope $m$ to $S$: $y=mx\pm\sqrt{25m^2-16}$.
   5. Tangent of slope $m$ to $C$: $y=mx\pm 4\sqrt{1+m^2}$.
   6. For a common tangent, the constants must match: $25m^2-16=16(1+m^2)$.
   7. $25m^2-16=16+16m^2\Rightarrow 9m^2=32\Rightarrow m^2=\frac{32}{9}\Rightarrow m=\pm\frac{4\sqrt2}{3}$.

   Common Mistakes

   • Taking the transverse axis of the conjugate hyperbola to have half-length $a$ instead of $b$.
   • Squaring the tangent-constant expressions incorrectly (must equate $c^2$ values, not $c$ values, since both are $\pm$).

   ✓Final answer

   The correct option is (B) — $\pm\frac{4\sqrt2}{3}$.

   ANSWER: B

7. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $d_1$ and $d_2$ are the distances of the foci of the hyperbola $4x^2 - 9y^2 - 16x + 54y - 101 = 0$ from the point (2 , −3), then $d_1 + d_2 =$ (A) 10 (B) 14 (C) 12 (D) 16

   ›Reveal solutionSolution

   This tests reducing a hyperbola to standard form and then computing plain distances from a given point to its two foci; the sum is 14.

   Concept and Intuition

   The point $(2,-3)$ is not on the hyperbola, so the constant-difference-of-distances property does not apply here — we simply need the ordinary distance from a point to each focus and then add them.

   Step-by-Step Solution

   1. Group and complete the square: $4x^2-16x-9y^2+54y-101=0$ $\Rightarrow 4(x^2-4x+4)-16-9(y^2-6y+9)+81-101=0$ $\Rightarrow 4(x-2)^2-9(y-3)^2-36=0 \Rightarrow 4(x-2)^2-9(y-3)^2=36$.
   2. Divide by 36: $\dfrac{(x-2)^2}{9}-\dfrac{(y-3)^2}{4}=1$. Centre $(2,3)$, $a^2=9,\,b^2=4$.
   3. $c^2=a^2+b^2=13\Rightarrow c=\sqrt{13}$. Foci: $(2+\sqrt{13},3)$ and $(2-\sqrt{13},3)$.
   4. The given point $(2,-3)$ has the same x-coordinate as the centre and is $6$ units below it.
   5. Distance to either focus $=\sqrt{(\sqrt{13})^2+6^2}=\sqrt{13+36}=\sqrt{49}=7$.
   6. $d_1+d_2 = 7+7 = 14$.

   Common Mistakes

   • Trying to use $|d_1-d_2|=2a$ even though $(2,-3)$ is not a point on the hyperbola.
   • Sign errors while completing the square.

   ✓Final answer

   The correct option is (B) — 14.

   ANSWER: B

8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If C is the centre of the hyperbola $\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1$ and the tangent drawn at any point P on the hyperbola meets the lines $bx - ay = 0$ and $bx + ay = 0$ at Q and R respectively, then $CQ \cdot CR =$ (A) $a^2 - b^2$ (B) $a^2 + b^2$ (C) $\dfrac{1}{a^2} + \dfrac{1}{b^2}$ (D) $\dfrac{1}{a^2} - \dfrac{1}{b^2}$

   ›Reveal solutionSolution

   Parametrising the hyperbola and intersecting its tangent with the two asymptotes gives

   $CQ\cdot CR = a^2+b^2$, independent of the point $P$ chosen — a clean invariant.

   Concept and Intuition

   The lines $bx\mp ay=0$ are exactly the asymptotes of the hyperbola $\tfrac{x^2}{a^2}-\tfrac{y^2}{b^2}=1$.

   A well-known property is that the tangent at any point of a hyperbola cuts its two asymptotes in

   points whose distances from the centre multiply to the constant $a^2+b^2$ — this is analogous to

   (and provable the same way as) the "tangent cuts a constant area triangle with the asymptotes"

   property.

   Step-by-Step Solution

   1. Let $P=(a\sec\theta, b\tan\theta)$. The tangent at $P$ is

   $\frac{x\sec\theta}{a}-\frac{y\tan\theta}{b}=1$

   2. Intersect with $y=\dfrac{b}{a}x$ (i.e. $bx-ay=0$):

   $\frac{x\sec\theta}{a}-\frac{\tan\theta}{b}\cdot\frac{b}{a}x=1 \Rightarrow \frac{x(\sec\theta-\tan\theta)}{a}=1 \Rightarrow x=\frac{a}{\sec\theta-\tan\theta}$

   Then $y=\dfrac{b}{\sec\theta-\tan\theta}$, so

   $CQ=\sqrt{x^2+y^2}=\frac{\sqrt{a^2+b^2}}{|\sec\theta-\tan\theta|}$

   3. Intersect with $y=-\dfrac{b}{a}x$ (i.e. $bx+ay=0$) similarly:

   $x=\frac{a}{\sec\theta+\tan\theta},\qquad CR=\frac{\sqrt{a^2+b^2}}{|\sec\theta+\tan\theta|}$

   4. Multiply:

   $CQ\cdot CR=\frac{a^2+b^2}{|(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)|} =\frac{a^2+b^2}{|\sec^2\theta-\tan^2\theta|}=\frac{a^2+b^2}{1}=a^2+b^2$

   (using $\sec^2\theta-\tan^2\theta=1$).

   Common Mistakes

   • Forgetting that $\sec^2\theta-\tan^2\theta=1$ simplifies the denominator to exactly $1$, making the whole product independent of $P$.
   • Sign confusion between the two asymptote lines when solving for $x$.

   ✓Final answer

   The correct option is (B) — $a^2 + b^2$.

   ANSWER: B

9. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.A normal is drawn to the hyperbola $9x^2 - 16y^2 = 144$ at one of the ends of its latus rectum. If that end lies in the third quadrant and the equation of the normal is $ax + by + c = 0$, then $\dfrac{b+c}{a} =$ (A) $\dfrac{44}{25}$ (B) $\dfrac{84}{25}$ (C) $\dfrac{55}{16}$ (D) $\dfrac{145}{16}$

   ›Reveal solutionSolution

   This tests locating the latus-rectum endpoints of a hyperbola in the correct quadrant and applying the standard normal-line formula at that point. The normal's coefficients give $\dfrac{b+c}{a}=\dfrac{145}{16}$.

   Concept and Intuition

   For $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, the foci are at $(\pm c,0)$ with $c^2=a^2+b^2$, and the latus rectum through each focus has half-length $b^2/a$, so its ends are $\left(\pm c,\pm\dfrac{b^2}a\right)$ (four combinations, one per quadrant containing a focus). The normal at a point $(x_1,y_1)$ on the hyperbola is obtained by differentiating implicitly to get the tangent slope, then taking the negative reciprocal for the normal slope — this simplifies to the clean symmetric form $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$, analogous to the ellipse's normal formula.

   Step-by-Step Solution

   1. $9x^2-16y^2=144$. Divide by $144$: $\dfrac{x^2}{16}-\dfrac{y^2}9=1$. So $a^2=16$ ($a=4$), $b^2=9$ ($b=3$), $c^2=a^2+b^2=25$ ($c=5$).
   2. Latus-rectum ends: $\left(\pm c,\pm\dfrac{b^2}{a}\right)=\left(\pm5,\pm\dfrac94\right)$.
   3. The end in the third quadrant (both coordinates negative) is $\left(-5,-\dfrac94\right)$.
   4. Normal to the hyperbola at $(x_1,y_1)$: $\dfrac{a^2x}{x_1}+\dfrac{b^2y}{y_1}=a^2+b^2$. With $a^2=16,b^2=9,x_1=-5,y_1=-\dfrac94$:

   $\frac{16x}{-5}+\frac{9y}{-9/4}=16+9=25.$

   5. Simplify $\dfrac{9y}{-9/4}=9y\cdot\left(-\dfrac49\right)=-4y$. So:

   $-\frac{16x}{5}-4y=25.$

   6. Multiply through by $-5$: $16x+20y=-125\Rightarrow16x+20y+125=0$.
   7. This matches the form $ax+by+c=0$ with $a=16,\,b=20,\,c=125$.
   8. $\frac{b+c}{a}=\frac{20+125}{16}=\frac{145}{16}.$ (This ratio is scale-invariant, so it doesn't matter that the coefficients $16,20,125$ aren't in lowest common-factor form as a triple.)

   Common Mistakes

   • Picking the latus-rectum end in the wrong quadrant (e.g. $(-5,9/4)$, second quadrant, instead of $(-5,-9/4)$, third quadrant).
   • Misremembering the hyperbola normal formula, confusing it with the ellipse's ($\dfrac{a^2x}{x_1}-\dfrac{b^2y}{y_1}=a^2-b^2$, which is different) — for a hyperbola both signs on the LHS stay positive relative to the equation's own sign convention, and the RHS is $a^2+b^2$.
   • Sign errors simplifying $\dfrac{9y}{-9/4}$.

   ✓Final answer

   The correct option is (D) — $\dfrac{145}{16}$.

   ANSWER: D

10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.One of the latus recta of the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2} = 1$ subtends an angle $2\mathrm{Tan}^{-1}\left(\dfrac{3}{2}\right)$ at the centre of the hyperbola. If $b^2 = 36$ and e is the eccentricity of the given hyperbola, then $\sqrt{a^2+e^2} =$ (A) $4$ (B) $\sqrt{14}$ (C) $6$ (D) $\sqrt{21}$

    ›Reveal solutionSolution

    Using the latus-rectum-subtended-angle formula $\tan\theta=\frac{b^2}{a^2e}$ together with the eccentricity relation $e^2=1+\frac{b^2}{a^2}$, solve for $a^2$ and $e$, then compute $\sqrt{a^2+e^2}$.

    Concept and Intuition

    A latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ passes through a focus $(\pm ae,0)$ and has half-length $\frac{b^2}{a}$, so its two endpoints are $(ae,\ b^2/a)$ and $(ae,\ -b^2/a)$. From the centre (origin), each endpoint subtends a half-angle $\theta$ with $\tan\theta=\frac{b^2/a}{ae}$, so the full angle between the two lines from the centre to the endpoints is $2\theta$. Combined with the standard hyperbola relation $b^2=a^2(e^2-1)$, this gives enough equations to solve for $a^2$ and $e$.

    Step-by-Step Solution

    1. Latus rectum endpoints: $(ae, b^2/a)$ and $(ae,-b^2/a)$. Half-angle at centre: $\tan\theta=\dfrac{b^2/a}{ae}=\dfrac{b^2}{a^2e}$.
    2. Given full angle $=2\tan^{-1}(3/2)$, so $\theta=\tan^{-1}(3/2) \Rightarrow \tan\theta=\frac32$.
    3. So $\dfrac{b^2}{a^2e}=\dfrac32$. With $b^2=36$: $\dfrac{36}{a^2e}=\dfrac32 \Rightarrow a^2e=24$.
    4. Eccentricity relation: $e^2=1+\dfrac{b^2}{a^2}=1+\dfrac{36}{a^2}$.
    5. Let $A=a^2$. From step 3, $e=\dfrac{24}{A}$, so $e^2=\dfrac{576}{A^2}$. Substitute into step 4:

    $\frac{576}{A^2}=1+\frac{36}{A}\ \Rightarrow\ 576=A^2+36A\ \Rightarrow\ A^2+36A-576=0.$

    6. Solve: $A=\dfrac{-36\pm\sqrt{1296+2304}}{2}=\dfrac{-36\pm60}{2}=12\text{ or }-48$. Since $A=a^2>0$, $a^2=12$.
    7. $e=\dfrac{24}{12}=2$. Check: $e^2=1+\frac{36}{12}=1+3=4=2^2$ ✓.
    8. $\sqrt{a^2+e^2}=\sqrt{12+4}=\sqrt{16}=4$.

    Common Mistakes

    • Using the wrong latus-rectum length (some remember it as $\frac{2b^2}{a}$, the full length, and mistakenly halve incorrectly when finding the endpoints' $y$-coordinate).
    • Forgetting the hyperbola's eccentricity relation is $e^2=1+b^2/a^2$ (not $e^2=1-b^2/a^2$, which is for the ellipse).

    ✓Final answer

    The correct option is (A) — $4$.

    ANSWER: A

11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If the equation of the hyperbola having (8, 3), (0, 3) as foci and $\dfrac{4}{3}$ as eccentricity is $\dfrac{(x-\alpha)^2}{p}-\dfrac{(y-\beta)^2}{q}=1$ then $p+q =$ (A) $\beta^2$ (B) $\alpha+\beta$ (C) $\alpha^2$ (D) $\alpha\beta$

    ›Reveal solutionSolution

    The centre of the hyperbola is the midpoint of its foci, giving $\alpha=4,\beta=3$; using $c=4$ and $e=4/3$ we find $a^2=9$, $b^2=7$, so $p+q=16$, which equals $\alpha^2$.

    Concept and Intuition

    A hyperbola centred away from the origin, $\frac{(x-\alpha)^2}{p}-\frac{(y-\beta)^2}{q}=1$, has its centre $(\alpha,\beta)$ at the midpoint of its two foci, and its focal parameters $c$ (half the distance between foci), $a=\sqrt p$ (semi-transverse axis) and $b=\sqrt q$ (semi-conjugate axis) obey $b^2=c^2-a^2$ and $e=c/a$, exactly as for a hyperbola centred at the origin — just shifted.

    Step-by-Step Solution

    1. Foci: $(8,3)$ and $(0,3)$ — same $y$-coordinate, so the transverse axis is horizontal, and the centre is their midpoint: $(\alpha,\beta)=\left(\frac{8+0}{2},3\right)=(4,3)$.
    2. Distance between foci $=2c=8 \Rightarrow c=4$.
    3. Eccentricity $e=\frac43=\frac{c}{a}\Rightarrow a=\frac{c}{e}=\frac{4}{4/3}=3\Rightarrow p=a^2=9$.
    4. $b^2=c^2-a^2=16-9=7\Rightarrow q=7$.
    5. $p+q=9+7=16$.
    6. Compare with $\alpha=4$: $\alpha^2=16=p+q$.

    Common Mistakes

    • Mixing up $\alpha$ (the $x$-coordinate of the centre) with $c$ (focal distance from centre) — they're numerically close here ($\alpha=4=c$) purely by coincidence of this problem's numbers, which can cause confusion when checking the answer.
    • Using the ellipse relation $b^2=a^2-c^2$ (sign flipped) instead of the hyperbola relation $b^2=c^2-a^2$.

    ✓Final answer

    The correct option is (C) — $\alpha^2$.

    ANSWER: C

12. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
    Q.If the equation of the tangent of the hyperbola $5x^2 - 9y^2 - 20x - 18y - 34 = 0$ which makes an angle $45°$ with the positive X-axis in positive direction is $x + by + c = 0$ then $b^2 + c^2 =$ (A) 2 or 13 (B) 5 or 26 (C) 2 or 26 (D) 26 or 28

    ›Reveal solutionSolution

    This tests the tangent condition for a hyperbola with given slope, after recentring by completing the square. The answer is $b^2+c^2=2$ or $26$.

    Concept and Intuition

    Any conic given by a general second-degree equation with no $xy$ term can be recentred by completing the square in $x$ and $y$ separately. Once in standard form $\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=1$, the classical tangent condition (line $Y=mX+c$ touches the hyperbola iff $c^2=a^2m^2-b^2$) applies directly in the shifted coordinates.

    Step-by-Step Solution

    1. Complete the square:

    $5x^2-20x = 5(x-2)^2-20,\qquad -9y^2-18y=-9(y+1)^2+9$

    So the equation becomes $5(x-2)^2-9(y+1)^2-45=0$, i.e.

    $\dfrac{(x-2)^2}{9}-\dfrac{(y+1)^2}{5}=1$

    Here $a^2=9,\ b^2=5$, and shifted coordinates are $X=x-2,\ Y=y+1$.

    2. A line making $45°$ with the positive X-axis has slope $m=\tan45°=1$.

    3. Tangent condition for $Y=mX+c$ to touch $\dfrac{X^2}{a^2}-\dfrac{Y^2}{b^2}=1$:

    $c^2=a^2m^2-b^2=9(1)^2-5=4 \implies c=\pm2$

    4. Convert back to $x,y$: $Y=X+c \Rightarrow (y+1)=(x-2)+c \Rightarrow x-y+(c-3)=0$.

    Matching to the given form $x+by+c=0$ (calling the constant $c'$ to avoid clash with the value above): $b=-1$, and

    • for $c=2$: $c'=2-3=-1$
    • for $c=-2$: $c'=-2-3=-5$

    5. Compute $b^2+c'^2$:

    $b^2+c'^2 = (-1)^2+(-1)^2=2 \quad\text{or}\quad (-1)^2+(-5)^2=1+25=26$

    Common Mistakes

    • Forgetting to complete the square first and applying the tangent formula to the un-centred equation.
    • Confusing the slope-form tangent condition for a hyperbola with that of an ellipse (sign of $b^2$ is different).
    • Losing the shift when converting the tangent line back from $X,Y$ to $x,y$.

    ✓Final answer

    The correct option is (C) — 2 or 26.

    ANSWER: C