Let H_1:fracx^2a^2-fracy^2b^2=1$H_{1}:\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ and H_2:-fracx^2A^2+fracy^2B^2=1$H_{2}:-\frac{x^{2}}{A^{2}}+\frac{y^{2}}{B^{2}}=1$ be two hyperbolas having length of latus rectums 15sqrt2$15\sqrt{2}$ and 12sqrt5$12\sqrt{5}$ respectively. Let their eccentricities be e_1=sqrtfrac52$e_{1}=\sqrt{\frac{5}{2}}$ and e_2$e_{2}$ respectively. If the product of the lengths of their transverse axes is 100sqrt10$100\sqrt{10}$ then 25e_2^2$25e_{2}^{2}$ is equal to \_\_\_\_. [cite: 3413, 3414, 3415, 3416, 3417, 3418]
Keywords:#hyperbola latus rectum transverse axis#JEE Main 2025 Evening Q74#conjugate hyperbola eccentricity parameters#product of geometric axes conic
More Conic Sections Previous-Year Questions — Page 3
Q732025Hyperbola
Consider the hyperbola fracx^2a^2 -fracy^2b^2 = 1$\frac{x^2}{a^2} -\frac{y^2}{b^2} = 1$ having one of its focus at mathrmP(-3,0)$\mathrm{P(-3,0)}$ . If the latus rectum through its other focus subtends a right angle at mathrmP$\mathrm{P}$ and a^2 b^2 = alpha sqrt2 -beta ,alpha ,beta in mathbbN$a^2 b^2 = \alpha \sqrt{2} -\beta ,\alpha ,\beta \in \mathbb{N}$ , calculate alpha + beta$\alpha + \beta$.
Numerical Answer.Answer: 1944 to 1944
Solution
### Related Formula
For a standard hyperbola:
- Focus positions are (pm ae, 0)$(\pm ae, 0)$.
- Length of semi-latus rectum is fracb^2a$\frac{b^2}{a}$.
- Eccentricity identity linkage: b^2 = a^2(e^2 - 1) implies a^2e^2 = a^2 + b^2$b^2 = a^2(e^2 - 1) \implies a^2e^2 = a^2 + b^2$.
### Core Logic
Given focus F_1 equiv (-ae, 0) equiv P(-3, 0)$F_1 \equiv (-ae, 0) \equiv P(-3, 0)$, so ae = 3$ae = 3$.
The other focus is F_2 equiv (ae, 0) equiv (3, 0)$F_2 \equiv (ae, 0) \equiv (3, 0)$.
The latus rectum passes vertically through F_2$F_2$, with endpoints L_1left(ae, fracb^2aright)$L_1\left(ae, \frac{b^2}{a}\right)$ and L_2left(ae, -fracb^2aright)$L_2\left(ae, -\frac{b^2}{a}\right)$.
This segment subtends a right angle at P(-ae, 0)$P(-ae, 0)$. By symmetry, the top half angle at P$P$ must be exactly 45^circ$45^\circ$.
### Step 1: Set Up Slope Relationship
Hyperbola diagram for Q73 - JEE Main 2025 Morning
Using the geometric slope relationship:
tan 45^circ = fractextheighttextbase = fracb^2/a2ae$\tan 45^\circ = \frac{\text{height}}{\text{base}} = \frac{b^2/a}{2ae}$1 = fracb^22a^2e implies 2a^2e = b^2 implies b^2 = 6a quad (textsince ae = 3)$1 = \frac{b^2}{2a^2e} \implies 2a^2e = b^2 \implies b^2 = 6a \quad (\text{since } ae = 3)$
### Step 2: Solve the Quadratic Excentricity Equation
Substitute ae = 3$ae = 3$ and b^2 = 6a$b^2 = 6a$ into the eccentricity identity a^2e^2 = a^2 + b^2$a^2e^2 = a^2 + b^2$:
9 = a^2 + 6a implies a^2 + 6a - 9 = 0$9 = a^2 + 6a \implies a^2 + 6a - 9 = 0$
Solving for a$a$ using the quadratic formula (taking the positive root since a > 0$a > 0$):
a = frac-6 pm sqrt36 - 4(1)(-9)2 = frac-6 + sqrt722 = -3 + 3sqrt2 = 3(sqrt2 - 1)$a = \frac{-6 \pm \sqrt{36 - 4(1)(-9)}}{2} = \frac{-6 + \sqrt{72}}{2} = -3 + 3\sqrt{2} = 3(\sqrt{2} - 1)$
### Step 3: Evaluate product and sum coefficients
Now compute a^2b^2$a^2b^2$:
a^2b^2 = a^2(6a) = 6a^3$a^2b^2 = a^2(6a) = 6a^3$6a^3 = 6left[3(sqrt2 - 1)right]^3 = 6 times 27 times (sqrt2 - 1)^3$6a^3 = 6\left[3(\sqrt{2} - 1)\right]^3 = 6 \times 27 \times (\sqrt{2} - 1)^3$6a^3 = 162 times (2sqrt2 - 6 + 3sqrt2 - 1) = 162 times (5sqrt2 - 7)$6a^3 = 162 \times (2\sqrt{2} - 6 + 3\sqrt{2} - 1) = 162 \times (5\sqrt{2} - 7)$6a^3 = 810sqrt2 - 1134$6a^3 = 810\sqrt{2} - 1134$
Matching with alphasqrt2 - beta$\alpha\sqrt{2} - \beta$ gives:
alpha = 810 quad textand quad beta = 1134$\alpha = 810 \quad \text{and} \quad \beta = 1134$
Calculate the final required sum:
alpha + beta = 810 + 1134 = 1944$\alpha + \beta = 810 + 1134 = 1944$
### Pattern Recognition
Recognizing that the right angle subtended at the opposite focus implies a perfect tan(45^circ)$\tan(45^\circ)$ right triangle instantly yields the key linear constraint b^2 = 2a(ae)$b^2 = 2a(ae)$, avoiding the need for lengthy distance-formula tracking.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q602025Ellipse and Focal Distances
Let the ellipse 3x^2 + py^2 = 4$3x^{2} + py^{2} = 4$ pass through the centre C of the circle x^2 + y^2 - 2x - 4y - 11 = 0$x^{2} + y^{2} - 2x - 4y - 11 = 0$ of radius r. Let f_1, f_2$f_{1}, f_{2}$ be the focal distances of the point C on the ellipse. Then 6f_1f_2 - r$6f_{1}f_{2} - r$ is equal to
A.74$74$
B.68$68$
C.70$70$
D.78$78$
Solution
### Related Formula
textFocal Distance Product on Vertical Ellipse = b^2 - e^2 k^2$\text{Focal Distance Product on Vertical Ellipse} = b^2 - e^2 k^2$
### Core Logic
Extract the coordinate center of the target circle, substitute it directly to locate the missing parameter p$p$, and resolve eccentricity metrics.
### Step 1: Extract Circle Metric Values
For circle x^2 + y^2 - 2x - 4y - 11 = 0$x^{2} + y^{2} - 2x - 4y - 11 = 0$:
textCentre C(1, 2), quad textRadius r = sqrt1 + 4 + 11 = 4$\text{Centre } C(1, 2), \quad \text{Radius } r = \sqrt{1 + 4 + 11} = 4$
### Step 2: Standardize Ellipse Formulation
Ellipse passes through point C(1,2)$C(1,2)$:
3(1)^2 + p(2)^2 = 4 implies 3 + 4p = 4 implies p = frac14$3(1)^2 + p(2)^2 = 4 \implies 3 + 4p = 4 \implies p = \frac{1}{4}$
Standard model form: fracx^24/3 + fracy^216 = 1$\frac{x^2}{4/3} + \frac{y^2}{16} = 1$ (b > a$b > a$, vertical configuration axis).
e = sqrt1 - frac4/316 = sqrt1 - frac112 = sqrtfrac1112$e = \sqrt{1 - \frac{4/3}{16}} = \sqrt{1 - \frac{1}{12}} = \sqrt{\frac{11}{12}}$
### Step 3: Evaluate Product Chain
Focal distance elements at ordinate coordinate height k=2$k=2$ are bounded by b pm ek$b \pm ek$:
f_1 f_2 = b^2 - e^2 k^2 = 16 - left(frac1112right) times 4 = 16 - frac113 = frac373$f_1 f_2 = b^2 - e^2 k^2 = 16 - \left(\frac{11}{12}\right) \times 4 = 16 - \frac{11}{3} = \frac{37}{3}$
Target evaluation expression response string:
6f_1 f_2 - r = 6 left(frac373right) - 4 = 74 - 4 = 70$6f_1 f_2 - r = 6 \left(\frac{37}{3}\right) - 4 = 74 - 4 = 70$
### Pattern Recognition
Pay attention to whether b > a$b > a$ or a > b$a > b$ when analyzing ellipse forms. Focal distance definitions swap directions immediately across major horizontal/vertical configurations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Circles
Q752025Tangent to Parabola and Circle Properties
Let r$r$ be the radius of the circle, which touches x$x$ -axis at point (a, 0)$(a, 0)$ , a < 0$a < 0$ and the parabola y^2 = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to
Numerical Answer.Answer: 30 to 30
Solution
### Related Formula
textTangent line at point (x_1, y_1) implies yy_1 = 2a(x+x_1)$\text{Tangent line at point } (x_1, y_1) \implies yy_1 = 2a(x+x_1)$
### Core Logic
Establish the tangent vector expression at the parabola intersection mark. Since this path line functions as a shared contact tangent boundaries sheet for the circular arc, impose radius equations.
### Step 1: Derive Shared Parabola Tangent Line
Tangent line profile for y^2 = 9x$y^2 = 9x$ at coordinate indicator (4,6)$(4,6)$:
6y = 9 cdot left( fracx+42 right) implies 3x - 4y + 12 = 0$6y = 9 \cdot \left( \frac{x+4}{2} \right) \implies 3x - 4y + 12 = 0$
### Step 2: Build Geometric Metric Connections
Circle touches axis at (a,0)$(a,0)$, mapping coordinates center directly to C(a,r)$C(a,r)$. Perpendicular boundary constraint steps require:
frac3a - 4r + 125 = pm r implies 3a + 12 = 4r pm 5r$\frac{3a - 4r + 12}{5} = \pm r \implies 3a + 12 = 4r \pm 5r$
### Step 3: Solve for Radius Matrix Bounds
Enforce circle equation intersection constraint profile (x-a)^2 + (y-r)^2 = r^2$(x-a)^2 + (y-r)^2 = r^2$ at point (4,6)$(4,6)$:
a^2 - 8a - 12r + 52 = 0$a^2 - 8a - 12r + 52 = 0$
Evaluating the target systems from structural logic tracks rejects positive value parameters, providing:
a = -14, quad r = 30$a = -14, \quad r = 30$
{{SOL_IMG_75}}
### Pattern Recognition
Shared tangent elements connect independent conic fields. Locating circular center boundaries using axial coordinate tracking simplifies secondary equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Circles
Q592025Chord 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
Q752025Properties 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
More Conic Sections Questions — jee_main_2025_24_jan_evening
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