Let the foci of hyperbola coincide with the foci of the ellipse fracx^236 +fracy^216 = 1$\frac{x^2}{36} +\frac{y^2}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:
A.12
B.16
C.frac96sqrt5$\frac{96}{\sqrt{5}}$
D.24sqrt5$24\sqrt{5}$
Solution & Explanation
### Related Formula
textEccentricity of ellipse e_1 = sqrt1 - fracb^2a^2$$\text{Eccentricity of ellipse } e_1 = \sqrt{1 - \frac{b^2}{a^2}}$$textFoci = (pm ae_1, 0)$$\text{Foci} = (\pm ae_1, 0)$$textLength of Latus Rectum of hyperbola = frac2b_hyp^2a_hyp$$\text{Length of Latus Rectum of hyperbola} = \frac{2b_{hyp}^2}{a_{hyp}}$$
### Core Logic
For the given ellipse fracx^236 + fracy^216 = 1$\frac{x^2}{36} + \frac{y^2}{16} = 1$:
a^2 = 36 Rightarrow a = 6$a^2 = 36 \Rightarrow a = 6$b^2 = 16$b^2 = 16$e_1 = sqrt1 - frac1636 = sqrt1 - frac49 = fracsqrt53$$e_1 = \sqrt{1 - \frac{16}{36}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$$
Foci of the ellipse are at (pm ae_1, 0) = left(pm 6 cdot fracsqrt53, 0right) = (pm 2sqrt5, 0)$(\pm ae_1, 0) = \left(\pm 6 \cdot \frac{\sqrt{5}}{3}, 0\right) = (\pm 2\sqrt{5}, 0)$.
### Step 1: Establish Hyperbola Parameters
Let the hyperbola be fracx^2p^2 - fracy^2q^2 = 1$\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$.
Its foci coincide with the ellipse, so the foci of hyperbola are also (pm 2sqrt5, 0)$(\pm 2\sqrt{5}, 0)$.
Let e$e$ be the eccentricity of the hyperbola. We are given e = 5$e = 5$.
Focus of hyperbola is pe = 2sqrt5$pe = 2\sqrt{5}$.
p(5) = 2sqrt5 Rightarrow p = frac2sqrt55 = frac2sqrt5$$p(5) = 2\sqrt{5} \Rightarrow p = \frac{2\sqrt{5}}{5} = \frac{2}{\sqrt{5}}$$
### Step 2: Find the Conjugate Axis (q)
For the hyperbola:
e^2 = 1 + fracq^2p^2$$e^2 = 1 + \frac{q^2}{p^2}$$25 = 1 + fracq^2left(frac2sqrt5right)^2$$25 = 1 + \frac{q^2}{\left(\frac{2}{\sqrt{5}}\right)^2}$$24 = fracq^24/5 Rightarrow 24 = frac5q^24$$24 = \frac{q^2}{4/5} \Rightarrow 24 = \frac{5q^2}{4}$$5q^2 = 96 Rightarrow q^2 = frac965$$5q^2 = 96 \Rightarrow q^2 = \frac{96}{5}$$
### Step 3: Calculate Latus Rectum
Length of Latus Rectum = frac2q^2p$\frac{2q^2}{p}$= frac2 left(frac965right)frac2sqrt5 = frac965 times sqrt5 = frac96sqrt5$$= \frac{2 \left(\frac{96}{5}\right)}{\frac{2}{\sqrt{5}}} = \frac{96}{5} \times \sqrt{5} = \frac{96}{\sqrt{5}}$$
### Pattern Recognition
Co-focal conics share the exact mathematical value of their focal length ae$ae$ (or pe$pe$). Instantly extract c = ae$c = ae$ from the first shape and map it directly to c = pe$c = pe$ for the second.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Maths: Conic Sections
Keywords:#foci of hyperbola coincide#JEE Main 2026 Morning Q11#Conic Sections JEE Main 2026#Coinciding Foci of Ellipse and Hyperbola JEE Main 2026
More Conic Sections Previous-Year Questions — Page 4
Q59jee_main_2025_29_jan_eveningChord with a Given Midpoint
If alpha x + beta y = 109$\alpha x + \beta y = 109$ is the equation of the chord of the ellipse fracx^29 +fracy^24 = 1$\frac{x^2}{9} +\frac{y^2}{4} = 1$, whose mid point is left(frac52,frac12right)$\left(\frac{5}{2},\frac{1}{2}\right)$, then alpha +beta$\alpha +\beta$ is equal to
A.37$37$
B.46$46$
C.58$58$
D.72$72$
Solution
### Related Formula
Equation of a chord of a conic section with a given midpoint (x_1, y_1)$(x_1, y_1)$ is:
T = S_1$T = S_1$
### Core Logic
Given midpoint Mleft(frac52, frac12
ight)$M\left(\frac{5}{2}, \frac{1}{2}
ight)$ and ellipse fracx^29 + fracy^24 = 1$\frac{x^2}{9} + \frac{y^2}{4} = 1$.
Chord with a Given Midpoint diagram for Q59 - JEE Main 2025 Evening
Write T$T$ and S_1$S_1$ terms:
T: fracxleft(frac52right)9 + fracyleft(frac12right)4$$T: \frac{x\left(\frac{5}{2}\right)}{9} + \frac{y\left(\frac{1}{2}\right)}{4}$$S_1: fracleft(frac52right)^29 + fracleft(frac12right)^24$$S_1: \frac{\left(\frac{5}{2}\right)^2}{9} + \frac{\left(\frac{1}{2}\right)^2}{4}$$
Equating both sides:
frac5x18 + fracy8 = frac2536 + frac116$$\frac{5x}{18} + \frac{y}{8} = \frac{25}{36} + \frac{1}{16}$$
### Step 1: Simplify to Standard Form
Multiply the entire equation by 144 to eliminate fractions:
144left(frac5x18right) + 144left(fracy8right) = 144left(frac2536right) + 144left(frac116right)$$144\left(\frac{5x}{18}\right) + 144\left(\frac{y}{8}\right) = 144\left(\frac{25}{36}\right) + 144\left(\frac{1}{16}\right)$$40x + 18y = 4(25) + 9(1)$$40x + 18y = 4(25) + 9(1)$$40x + 18y = 109$40x + 18y = 109$
Comparing this directly with alpha x + beta y = 109$\alpha x + \beta y = 109$ provides:
alpha = 40, quad beta = 18$$\alpha = 40, \quad \beta = 18$$alpha + beta = 40 + 18 = 58$$\alpha + \beta = 40 + 18 = 58$$
### Pattern Recognition
Whenever you see 'chord whose midpoint is given', write T = S_1$T = S_1$ automatically. Match coefficients directly at the final step after equating constant integers.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q75jee_main_2025_29_jan_eveningProperties of Focal Chords
Let y^2 = 12x$y^2 = 12x$ the parabola and S$S$ be its focus. Let PQ$PQ$ be a focal chord of the parabola such that (SP) (SQ) = frac1474$(SQ) = \frac{147}{4}$. Let C$C$ be the circle described taking PQ$PQ$ as a diameter. If the equation of a circle C$C$ is 64x^2 + 64y^2 - alpha x - 64sqrt3y = beta$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$, then \beta - \alpha is equal to
Numerical Answer.Answer: 1328 to 1328
Solution
### Related Formula
Properties of focal chord parameter metrics in parabolas y^2 = 4ax$y^2 = 4ax$:
t_1 cdot t_2 = -1$$t_1 \cdot t_2 = -1$$
Distance to the directrix property:
SP = a(1 + t^2), quad SQ = aleft(1 + frac1t^2right)$$SP = a(1 + t^2), \quad SQ = a\left(1 + \frac{1}{t^2}\right)$$
### Core Logic
Given parabola y^2 = 12x implies a = 3$y^2 = 12x \implies a = 3$. Focus S = (3, 0)$S = (3, 0)$.
Set up focal segments product equation:
SP cdot SQ = 3(1+t^2) cdot 3left(1+frac1t^2right) = frac1474$$SP \cdot SQ = 3(1+t^2) \cdot 3\left(1+\frac{1}{t^2}\right) = \frac{147}{4}$$9 cdot frac(1+t^2)^2t^2 = frac1474 implies frac(1+t^2)^2t^2 = frac4912$$9 \cdot \frac{(1+t^2)^2}{t^2} = \frac{147}{4} \implies \frac{(1+t^2)^2}{t^2} = \frac{49}{12}$$
Solving for t^2$t^2$:
12t^4 - 25t^2 + 12 = 0 implies t^2 = frac34 quad textor quad frac43$$12t^4 - 25t^2 + 12 = 0 \implies t^2 = \frac{3}{4} \quad \text{or} \quad \frac{4}{3}$$
### Step 1: Compute Endpoint Coordinate Bounds
Choosing t = -fracsqrt32$t = -\frac{\sqrt{3}}{2}$ allows defining both chord coordinates symmetrically:
P(3t^2, 6t) implies Pleft(frac94, -3sqrt3right)$$P(3t^2, 6t) \implies P\left(\frac{9}{4}, -3\sqrt{3}\right)$$Qleft(frac3t^2, -frac6tright) implies Q(4, 4sqrt3)$$Q\left(\frac{3}{t^2}, -\frac{6}{t}\right) \implies Q(4, 4\sqrt{3})$$
### Step 2: Derive Circle Equation
Write the diameter circle form equation:
(x - 4)left(x - frac94right) + (y - 4sqrt3)(y + 3sqrt3) = 0$$(x - 4)\left(x - \frac{9}{4}\right) + (y - 4\sqrt{3})(y + 3\sqrt{3}) = 0$$x^2 + y^2 - frac254x - sqrt3y - 27 = 0$$x^2 + y^2 - \frac{25}{4}x - \sqrt{3}y - 27 = 0$$
Multiply by 64 to clear the fractions and match the given equation template structure:
64x^2 + 64y^2 - 400x - 64sqrt3y - 1728 = 0$$64x^2 + 64y^2 - 400x - 64\sqrt{3}y - 1728 = 0$$
Comparing directly with 64x^2 + 64y^2 - alpha x - 64sqrt3y = beta$64x^2 + 64y^2 - \alpha x - 64\sqrt{3}y = \beta$ yields:
alpha = 400, quad beta = 1728$$\alpha = 400, \quad \beta = 1728$$beta - alpha = 1728 - 400 = 1328$$\beta - \alpha = 1728 - 400 = 1328$$
### Pattern Recognition
The distance from focal chord endpoints to the focus equals their perpendicular distance to the directrix. This property connects parameter metrics to geometric lengths cleanly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Circles
Q52jee_main_2025_28_jan_morningParabola and Trapezium Properties
Let ABCD be a trapezium whose vertices lie on the parabolay^2 = 4x$y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length frac254$\frac{25}{4}$ and it passes through the point (1,0)$(1,0)$, then the area of ABCD is:
(1) frac754$\frac{75}{4}$
(2) frac252$\frac{25}{2}$
(3) frac1258$\frac{125}{8}$
(4) frac758$\frac{75}{8}$
A.frac754$\frac{75}{4}$
B.frac252$\frac{25}{2}$
C.frac1258$\frac{125}{8}$
D.frac758$\frac{75}{8}$
Solution
### Related Formula
Area of a trapezium is given by:
textArea = frac12 times (textsum of parallel sides) times (textdistance between them)$$\text{Area} = \frac{1}{2} \times (\text{sum of parallel sides}) \times (\text{distance between them})$$
### Core Logic
Let the coordinates of the vertices be parameterized on the parabola y^2 = 4x$y^2 = 4x$. Since AD$AD$ and BC$BC$ are parallel to the y-axis, the coordinates take the form:
A(at_1^2, 2at_1)$A(at_1^2, 2at_1)$ and D(at_1^2, -2at_1)$D(at_1^2, -2at_1)$B(at_2^2, 2at_2)$B(at_2^2, 2at_2)$ and C(at_2^2, -2at_2)$C(at_2^2, -2at_2)$
Given a=1$a=1$, the points simplify accordingly. Parabola and Trapezium Properties diagram for Q52 - JEE Main 2025 Morning
### Step 1: Using Diagonal Properties
The length of diagonal AC$AC$ passing through focal point (1,0)$(1,0)$ implies focal chord properties:
textLength AC = aleft(t_1 + frac1t_1right)^2 = frac254$$\text{Length AC} = a\left(t_1 + \frac{1}{t_1}\right)^2 = \frac{25}{4}$$t_1 + frac1t_1 = pmfrac52 implies t_1 = 2 text or frac12$$t_1 + \frac{1}{t_1} = \pm\frac{5}{2} \implies t_1 = 2 \text{ or } \frac{1}{2}$$
### Step 2: Finding Coordinates and Area
Substituting t_1 = 2$t_1 = 2$, we get:
Aleft(frac12, 1right), Dleft(frac14, -1right), B(4, 4), C(4, -4)$A\left(\frac{1}{2}, 1\right), D\left(\frac{1}{4}, -1\right), B(4, 4), C(4, -4)$
Evaluating the area formula:
textArea = frac12 times (8 + 2) times left(4 - frac14
ight) = frac754$$\text{Area} = \frac{1}{2} \times (8 + 2) \times \left(4 - \frac{1}{4}
ight) = \frac{75}{4}$$
### Pattern Recognition
Focal chords of parabolas always satisfy t_1 t_2 = -1$t_1 t_2 = -1$. Recognizing the passage through (1,0)$(1,0)$ unlocks quick parametric simplifications.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Conic Sections
Q75jee_main_2025_28_jan_morningInfinite Series of Ellipses
Let E_1: fracx^29 + fracy^24 = 1$E_1: \frac{x^2}{9} + \frac{y^2}{4} = 1$ be an ellipse. Ellipses E_i$E_i$ 's are constructed such that their centres and eccentricities are same as that of E_1$E_1$ , and the length of minor axis of E_i$E_i$ is the length of major axis of E_i+1$E_{i+1}$ ( i ge 1$i \ge 1$ ). If A_i$A_i$ is the area of the ellipse E_i$E_i$ , then frac5pi left( sum_i=1^infty A_i right)$\frac{5}{pi} \left( \sum_{i=1}^{\infty} A_i \right)$ , is equal to ....
Numerical Answer.Answer: 54 to 54
Solution
### Related Formula
Area of an ellipse with semi-axes a$a$ and b$b$:
textArea = pi a b$$\text{Area} = \pi a b$$
### Core Logic
Calculate the constant eccentricity e$e$ from the initial ellipse E_1$E_1$: Infinite Series of Ellipses diagram for Q75 - JEE Main 2025 Morninge = sqrt1 - frac49 = fracsqrt53$$e = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3}$$
For any subsequent ellipse E_2$E_2$, its major axis equals the minor axis of E_1$E_1$ (2b_1 = 4 implies a_2 = 2$2b_1 = 4 \implies a_2 = 2$). Since eccentricity remains constant:
frac59 = 1 - fracb_2^2a_2^2 = 1 - fracb_2^24 implies b_2^2 = frac169 implies b_2 = frac43$$\frac{5}{9} = 1 - \frac{b_2^2}{a_2^2} = 1 - \frac{b_2^2}{4} \implies b_2^2 = \frac{16}{9} \implies b_2 = \frac{4}{3}$$
### Step 1: Finding the Area Sequence Terms
Evaluate the area values for the initial ellipses:
A_1 = pi cdot 3 cdot 2 = 6pi$A_1 = \pi \cdot 3 \cdot 2 = 6\pi$A_2 = pi cdot 2 cdot frac43 = frac8pi3$A_2 = \pi \cdot 2 \cdot \frac{4}{3} = \frac{8\pi}{3}$
The areas form an infinite geometric progression with a common ratio r = frac49$r = \frac{4}{9}$.
### Step 2: Summing the Infinite Geometric Series
sum_i=1^infty A_i = frac6pi1 - frac49 = frac6pifrac59 = frac54pi5$$\sum_{i=1}^{\infty} A_i = \frac{6\pi}{1 - \frac{4}{9}} = \frac{6\pi}{\frac{5}{9}} = \frac{54\pi}{5}$$
Evaluating the final scaling formula:
frac5pi left( frac54pi5 right) = 54$$\frac{5}{\pi} \left( \frac{54\pi}{5} \right) = 54$$
### Pattern Recognition
Iterative dimensional scaling creates geometric progressions where the ratio equals the square of the linear scaling factor.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Conic Sections
Q56jee_main_2025_03_april_morningEllipse and Line Properties
A line passing through the point P(sqrt5, sqrt5)$P(\sqrt{5}, \sqrt{5})$ intersects the ellipse fracx^236 + fracy^225 = 1$\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A$A$ and B$B$ [cite: 567] such that (PA) cdot (PB)$(PA) \cdot (PB)$ is maximum. Then 5(PA^2 + PB^2)$5(PA^{2} + PB^{2})$ is equal to[cite: 570]:
A. 218
B. 377
C. 290
D. 338
Solution
### Related Formula
Parametric line equation relative to an offset point P(x_0, y_0)$P(x_0, y_0)$:
x = x_0 + rcostheta, quad y = y_0 + rsintheta$$x = x_0 + r\cos\theta, \quad y = y_0 + r\sin\theta$$Ellipse and Line Properties diagram for Q56 - JEE Main 2025 Morning
### Core Logic
Assume any line through P$P$ can be represented parametrically by [cite: 1277]:
Q(sqrt5 + rcostheta, sqrt5 + rsintheta)$$Q(\sqrt{5} + r\cos\theta, \sqrt{5} + r\sin\theta)$$ [cite: 1277]
Substitute coordinates into the standard ellipse formula [cite: 1278]:
25(sqrt5 + rcostheta)^2 + 36(sqrt5 + rsintheta)^2 = 900$$25(\sqrt{5} + r\cos\theta)^2 + 36(sqrt{5} + r\sin\theta)^2 = 900$$ [cite: 1278]
Expanding and gathering powers of r$r$ yields [cite: 1280]:
r^2(25cos^2theta + 36sin^2theta) + 2sqrt5r(25costheta + 36sintheta) - 595 = 0$$r^2(25\cos^2\theta + 36\sin^2\theta) + 2\sqrt{5}r(25\cos\theta + 36\sin\theta) - 595 = 0$$ [cite: 1280]
The product of roots corresponds to the distance product value[cite: 1281, 1283]:
PA cdot PB = |r_1 r_2| = frac59525cos^2theta + 36sin^2theta = frac59525 + 11sin^2theta$$PA \cdot PB = |r_1 r_2| = \frac{595}{25\cos^2\theta + 36\sin^2\theta} = \frac{595}{25 + 11\sin^2\theta}$$ [cite: 1283]
### Step 1: Maximization condition
To make PA cdot PB$PA \cdot PB$ maximum, the denominator must be minimized [cite: 1284]:
sin^2theta = 0 implies theta = 0$$\sin^2\theta = 0 \implies \theta = 0$$ [cite: 1284]
This implies the chord line AB$AB$ must run parallel to the x-axis [cite: 1285]:
y_A = y_B = sqrt5$$y_A = y_B = \sqrt{5}$$ [cite: 1285]
Substitute y = sqrt5$y = \sqrt{5}$ back into the ellipse equation to calculate x-coordinates [cite: 1286]:
fracx^236 + frac525 = 1 implies fracx^236 = frac45 implies x^2 = frac1445$$\frac{x^2}{36} + \frac{5}{25} = 1 \implies \frac{x^2}{36} = \frac{4}{5} \implies x^2 = \frac{144}{5}$$ [cite: 1287]
Therefore, the coordinates are x = pm frac12sqrt5$x = \pm \frac{12}{\sqrt{5}}$.
### Step 2: Distance value summation
Compute PA^2 + PB^2$PA^2 + PB^2$ using coordinates directly [cite: 1289]:
PA^2 + PB^2 = left(sqrt5 - frac12sqrt5right)^2 + left(sqrt5 + frac12sqrt5right)^2$$PA^2 + PB^2 = \left(\sqrt{5} - \frac{12}{\sqrt{5}}\right)^2 + \left(\sqrt{5} + \frac{12}{\sqrt{5}}\right)^2$$ [cite: 1289]
= 2left(5 + frac1445right) = frac3385$$= 2\left(5 + \frac{144}{5}\right) = \frac{338}{5}$$ [cite: 1290]
Multiplying by 5 gives the target integer answer [cite: 1290]:
5(PA^2 + PB^2) = 338$$5(PA^2 + PB^2) = 338$$ [cite: 1290]
### Pattern Recognition
Parametric distances from a point intersecting a conic configuration usually form a standard quadratic equation in r$r$. The angle parameter immediately simplifies the boundary constraint optimization.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections (Ellipse)
More Conic Sections Questions — jee_main_2026_21_jan_morning
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