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
null
### 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 3
Q75jee_main_2025_03_april_eveningHyperbola
If the equation of the hyperbola with foci (4, 2)$(4, 2)$ and (8, 2)$(8, 2)$ is 3x^2 - y^2 - alpha x + beta y + gamma = 0$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$, then alpha + beta + gamma$\alpha + \beta + \gamma$ is equal to
Numerical Answer.Answer: 141 to 141
Solution
### Related Formula
For a horizontal hyperbola centered at (h,k)$(h,k)$:
frac(x-h)^2a^2 - frac(y-k)^2b^2 = 1$$\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$$
- Foci: (h pm ae, k)$(h \pm ae, k)$
- Eccentricity relation: b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$
### Core Logic
Foci are S_1 = (4,2)$S_1 = (4,2)$ and S_2 = (8,2)$S_2 = (8,2)$.
- Center C(h,k)$C(h,k)$ is the midpoint:
h = frac4 + 82 = 6, quad k = 2 implies C = (6, 2)$$h = \frac{4 + 8}{2} = 6, \quad k = 2 \implies C = (6, 2)$$
- Distance between foci:
2ae = 8 - 4 = 4 implies ae = 2$$2ae = 8 - 4 = 4 \implies ae = 2$$
Thus, b^2 = 4 - a^2$b^2 = 4 - a^2$.
### Step 1: Expanding standard equation
The equation is:
frac(x-6)^2a^2 - frac(y-2)^24-a^2 = 1$$\frac{(x-6)^2}{a^2} - \frac{(y-2)^2}{4-a^2} = 1$$(4-a^2)(x-6)^2 - a^2(y-2)^2 = a^2(4-a^2)$$(4-a^2)(x-6)^2 - a^2(y-2)^2 = a^2(4-a^2)$$
Comparing with 3x^2 - y^2 - alpha x + beta y + gamma = 0$3x^2 - y^2 - \alpha x + \beta y + \gamma = 0$, the ratio of coefficients of x^2$x^2$ and y^2$y^2$ is frac3-1 = -3$\frac{3}{-1} = -3$:
frac4 - a^2-a^2 = -3 implies 4 - a^2 = 3a^2 implies 4a^2 = 4 implies a^2 = 1$$\frac{4 - a^2}{-a^2} = -3 \implies 4 - a^2 = 3a^2 \implies 4a^2 = 4 \implies a^2 = 1$$
Thus, b^2 = 4 - 1 = 3$b^2 = 4 - 1 = 3$.
### Step 2: Finding values of coefficients alpha, beta, gamma$\alpha, \beta, \gamma$
Substituting a^2 = 1$a^2 = 1$ back into standard form equation:
3(x-6)^2 - (y-2)^2 = 3$$3(x-6)^2 - (y-2)^2 = 3$$3(x^2 - 12x + 36) - (y^2 - 4y + 4) = 3$$3(x^2 - 12x + 36) - (y^2 - 4y + 4) = 3$$3x^2 - 36x + 108 - y^2 + 4y - 4 = 3$$3x^2 - 36x + 108 - y^2 + 4y - 4 = 3$$3x^2 - y^2 - 36x + 4y + 101 = 0$$3x^2 - y^2 - 36x + 4y + 101 = 0$$
Comparing coefficients:
- alpha = 36$\alpha = 36$
- beta = 4$\beta = 4$
- gamma = 101$\gamma = 101$alpha + beta + gamma = 36 + 4 + 101 = 141$$\alpha + \beta + \gamma = 36 + 4 + 101 = 141$$Hyperbola diagram for Q75 - JEE Main 2025 Evening Shift
### Pattern Recognition
Symmetric focal coordinates (y=2$y=2$) indicate the hyperbola is horizontal. Identifying coordinates of the center (6,2)$(6,2)$ quickly and using coefficient ratio comparison restricts parameters immediately without requiring complex algebraic systems.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Conic Sections
Qjee_main_2025_07_april_morningParabola
Let mathrmP$\mathrm{P}$ be the parabola, whose focus is (-2, 1)$(-2, 1)$ and directrix is 2mathrmx + mathrmy + 2 = 0$2\mathrm{x} + \mathrm{y} + 2 = 0$ . Then the sum of the ordinates of the points on mathrmP$\mathrm{P}$ , whose abscissa is -2$-2$ , is
A.frac32$\frac{3}{2}$
B.frac52$\frac{5}{2}$
C.frac14$\frac{1}{4}$
D.frac34$\frac{3}{4}$
Solution
### Related Formula
By the definition of a parabola, the distance from any point (x, y)$(x, y)$ on the curve to the focus (x_f, y_f)$(x_f, y_f)$ equals its perpendicular distance to the directrix line Ax + By + C = 0$Ax + By + C = 0$:
(x - x_f)^2 + (y - y_f)^2 = frac(Ax + By + C)^2A^2 + B^2$$(x - x_f)^2 + (y - y_f)^2 = \frac{(Ax + By + C)^2}{A^2 + B^2}$$
### Core Logic
Substituting the focus (-2, 1)$(-2, 1)$ and directrix 2x + y + 2 = 0$2x + y + 2 = 0$ into the definition equation:
(x + 2)^2 + (y - 1)^2 = frac(2x + y + 2)^22^2 + 1^2$$(x + 2)^2 + (y - 1)^2 = \frac{(2x + y + 2)^2}{2^2 + 1^2}$$5left[(x + 2)^2 + (y - 1)^2right] = (2x + y + 2)^2$$5\left[(x + 2)^2 + (y - 1)^2\right] = (2x + y + 2)^2$$
### Step 1: Substitute the Given Abscissa
Parabola diagram for Q55 - JEE Main 2025 Morning
We need the points whose abscissa (x-coordinate) is x = -2$x = -2$. Substitute x = -2$x = -2$ into the general equation:
5left[(-2 + 2)^2 + (y - 1)^2right] = (2(-2) + y + 2)^2$$5\left[(-2 + 2)^2 + (y - 1)^2\right] = (2(-2) + y + 2)^2$$5left[0 + (y - 1)^2right] = (-4 + y + 2)^2$$5\left[0 + (y - 1)^2\right] = (-4 + y + 2)^2$$5(y - 1)^2 = (y - 2)^2$$5(y - 1)^2 = (y - 2)^2$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$5(y^2 - 2y + 1) = y^2 - 4y + 4$$5y^2 - 10y + 5 = y^2 - 4y + 4$$5y^2 - 10y + 5 = y^2 - 4y + 4$$4y^2 - 6y + 1 = 0$$4y^2 - 6y + 1 = 0$$
### Step 2: Find the Sum of Ordinates
The ordinates y_1$y_1$ and y_2$y_2$ are the roots of the quadratic equation 4y^2 - 6y + 1 = 0$4y^2 - 6y + 1 = 0$.
The sum of the ordinates is:
y_1 + y_2 = -frac-64 = frac64 = frac32$$y_1 + y_2 = -\frac{-6}{4} = \frac{6}{4} = \frac{3}{2}$$
### Pattern Recognition
Notice how evaluating the intersection layout directly simplifies when the substitution value matches the coordinate of the focus, converting the entire quadratic horizontal layout component to 0 immediately.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q73jee_main_2025_07_april_morningHyperbola
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
Q60jee_main_2025_08_april_eveningEllipse 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
Q75jee_main_2025_08_april_eveningTangent 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
More Conic Sections Questions — jee_main_2026_21_jan_morning
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