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
Q20jee_main_2026_21_jan_morningLocus of Internal Section Point
Let O be the vertex of the parabola x^2=4y$x^{2}=4y$ and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio 2: 3 be the conic C. Then the equation of the chord of C, which is bisected at the point (1, 2), is:
A. 5x-y-3=0
B. 4x-5y+6=0
C. x-2y + 3 = 0
D. 5x-4y+3=0
Solution
### Related Formula
textSection Formula: quad P = fracm cdot Q + n cdot Om + n$$\text{Section Formula:} \quad P = \frac{m \cdot Q + n \cdot O}{m + n}$$textChord bisected at (x_1, y_1) : quad T = S_1$$\text{Chord bisected at } (x_1, y_1) : \quad T = S_1$$
### Core Logic
Given parabola x^2 = 4y$x^2 = 4y$, its vertex O = (0, 0)$O = (0, 0)$.
A general point Q$Q$ on x^2 = 4y$x^2 = 4y$ is (2t, t^2)$(2t, t^2)$.
Let P(h, k)$P(h, k)$ divide OQ$OQ$ in ratio 2:3$2:3$.
By section formula:
h = frac2(2t) + 3(0)5 = frac4t5$$h = \frac{2(2t) + 3(0)}{5} = \frac{4t}{5}$$k = frac2(t^2) + 3(0)5 = frac2t^25$$k = \frac{2(t^2) + 3(0)}{5} = \frac{2t^2}{5}$$Locus of point dividing parabolic chord for Q20 - JEE Main 2026 Morning
### Step 1: Finding the Locus C
From h = frac4t5$h = \frac{4t}{5}$, we get t = frac5h4$t = \frac{5h}{4}$.
Substitute into k$k$:
k = frac25 left(frac5h4right)^2 = frac25 cdot frac25h^216 = frac5h^28$$k = \frac{2}{5} \left(\frac{5h}{4}\right)^2 = \frac{2}{5} \cdot \frac{25h^2}{16} = \frac{5h^2}{8}$$8k = 5h^2 Rightarrow 5x^2 = 8y$$8k = 5h^2 \Rightarrow 5x^2 = 8y$$
So the conic C is the parabola 5x^2 = 8y$5x^2 = 8y$.
### Step 2: Chord bisected at a point
We need the equation of the chord of C: 5x^2 - 8y = 0$C: 5x^2 - 8y = 0$ bisected at (x_1, y_1) = (1, 2)$(x_1, y_1) = (1, 2)$.
Use T = S_1$T = S_1$.
T = 5xx_1 - 4(y + y_1) = 5x(1) - 4(y + 2) = 5x - 4y - 8$T = 5xx_1 - 4(y + y_1) = 5x(1) - 4(y + 2) = 5x - 4y - 8$S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$
Equating T$T$ and S_1$S_1$:
5x - 4y - 8 = -11$$5x - 4y - 8 = -11$$5x - 4y + 3 = 0$5x - 4y + 3 = 0$
### Pattern Recognition
Internal division locus of a vertex chord on standard conic identically scales the conic. Once the child-conic is found, standard mid-point chord protocol (T=S_1$T=S_1$) strictly applies algebraically.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Maths: Parabola
Class 11 Maths: Straight Lines
Q55jee_main_2025_02_april_eveningEllipse
If the length of the minor axis of an ellipse is equal to one fourth of the distance between the foci, then the eccentricity of the ellipse is :
A.frac4sqrt17$\frac{4}{\sqrt{17}}$
B.fracsqrt316$\frac{\sqrt{3}}{16}$
C.frac3sqrt19$\frac{3}{\sqrt{19}}$
D.fracsqrt57$\frac{\sqrt{5}}{7}$
Solution
### Related Formula
textLength of minor axis = 2b$$\text{Length of minor axis} = 2b$$textDistance between foci = 2ae$$\text{Distance between foci} = 2ae$$textEccentricity: e = sqrt1 - fracb^2a^2$$\text{Eccentricity: } e = \sqrt{1 - \frac{b^2}{a^2}}$$
### Core Logic
We set up an algebraic equation relating b$b$, a$a$, and e$e$ from the given geometric condition, then substitute it into the eccentricity identity.
### Step 1: Set up the geometric relation
Given that 2b = frac14 (2ae)$2b = \frac{1}{4} (2ae)$:
b = fracae4 implies fracba = frace4$$b = \frac{ae}{4} \implies \frac{b}{a} = \frac{e}{4}$$
Square both sides:
fracb^2a^2 = frace^216$$\frac{b^2}{a^2} = \frac{e^2}{16}$$
### Step 2: Solve for eccentricity
Using the eccentricity relation:
e^2 = 1 - fracb^2a^2$$e^2 = 1 - \frac{b^2}{a^2}$$e^2 = 1 - frace^216$$e^2 = 1 - \frac{e^2}{16}$$e^2 left(1 + frac116right) = 1$$e^2 \left(1 + \frac{1}{16}\right) = 1$$frac1716 e^2 = 1 implies e^2 = frac1617 implies e = frac4sqrt17$$\frac{17}{16} e^2 = 1 \implies e^2 = \frac{16}{17} \implies e = \frac{4}{\sqrt{17}}$$
### Pattern Recognition
Standard Ellipse relations: For standard ellipses, the ratio of axes and the eccentricity are coupled quadratic equations. Expressing b/a$b/a$ as a function of e$e$ allows direct solving of the eccentricity.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q67jee_main_2025_02_april_eveningParabola
Let the point mathrmP$\mathrm{P}$ of the focal chord mathrmPQ$\mathrm{PQ}$ of the parabola mathrmy^2 = 16mathrmx$\mathrm{y}^2 = 16\mathrm{x}$ be (1, -4)$(1, -4)$. If the focus of the parabola divides the chord mathrmPQ$\mathrm{PQ}$ in the ratio mathrmm : mathrmn$\mathrm{m} : \mathrm{n}$, gcd(mathrmm, mathrmn) = 1$\gcd(\mathrm{m}, \mathrm{n}) = 1$, then mathrmm^2 + mathrmn^2$\mathrm{m}^2 + \mathrm{n}^2$ is equal to:
A. 17
B. 10
C. 37
D. 26
Solution
### Related Formula
textParametric coordinates on y^2 = 4ax: quad (at^2, \, 2at)$$\text{Parametric coordinates on } y^2 = 4ax: \quad (at^2, \, 2at)$$textFocal Chord relation: t_1 t_2 = -1$$\text{Focal Chord relation: } t_1 t_2 = -1$$textSection Formula: (x_c, y_c) = left( fracm x_2 + n x_1m+n, \, fracm y_2 + n y_1m+n right)$$\text{Section Formula: } (x_c, y_c) = \left( \frac{m x_2 + n x_1}{m+n}, \, \frac{m y_2 + n y_1}{m+n} \right)$$
### Core Logic
We find the parametric parameters of coordinates P$P$ and Q$Q$, obtain their Cartesian values, and then apply the section formula with the focus S$S$ to calculate the splitting ratio.
### Step 1: Find coordinates of P and Q
For parabola y^2 = 16x$y^2 = 16x$, the focal parameter is a = 4$a = 4$.
Focus is S(4, 0)$S(4, 0)$.
Let P$P$ be (a t_1^2, 2a t_1) = (1, -4)$(a t_1^2, 2a t_1) = (1, -4)$:
2a t_1 = -4 implies 2(4) t_1 = -4 implies t_1 = -frac12$$2a t_1 = -4 \implies 2(4) t_1 = -4 \implies t_1 = -\frac{1}{2}$$
Since PQ$PQ$ is a focal chord, the parametric points are coupled:
t_1 t_2 = -1 implies t_2 = 2$$t_1 t_2 = -1 \implies t_2 = 2$$
Now, calculate the coordinates of Q$Q$:
Q equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$$Q \equiv (a t_2^2, \, 2 a t_2) = (4(4), \, 2(4)(2)) = (16, \, 16)$$
### Step 2: Solve for the dividing ratio
Let the focus S(4, 0)$S(4, 0)$ divide the line segment PQ$PQ$ internally in the ratio lambda : 1$\lambda : 1$.
Using the y$y$-coordinate of the section formula:
y_s = fraclambda y_q + 1 y_plambda + 1$$y_s = \frac{\lambda y_q + 1 y_p}{\lambda + 1}$$0 = fraclambda(16) + 1(-4)lambda + 1 implies 16lambda - 4 = 0 implies lambda = frac14$$0 = \frac{\lambda(16) + 1(-4)}{\lambda + 1} \implies 16\lambda - 4 = 0 \implies \lambda = \frac{1}{4}$$
Thus, the focus S$S$ divides the chord internally in the ratio 1:4$1:4$.
Since gcd(1, 4) = 1$\gcd(1, 4) = 1$, we have m = 1$m = 1$ and n = 4$n = 4$:
m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$$m^2 + n^2 = 1^2 + 4^2 = 1 + 16 = 17$$
### Pattern Recognition
Harmonic Mean Shortcut: In any parabola, the focus divides a focal chord internally into segments of lengths SP$SP$ and SQ$SQ$ such that the semi-latus rectum 2a$2a$ is the harmonic mean of these segments: frac1SP + frac1SQ = frac1a$\frac{1}{SP} + \frac{1}{SQ} = \frac{1}{a}$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Qjee_main_2025_02_april_morningProperties of Hyperbola
Let one focus of the hyperbola H: fracx^2a^2 - fracy^2b^2 = 1$H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be at (sqrt10, 0)$(\sqrt{10}, 0)$ and the corresponding directrix be x = frac9sqrt10$x = \frac{9}{\sqrt{10}}$. If e$e$ and l$l$ respectively are the eccentricity and the length of the latus rectum of H$H$, then 9(e^2 + l)$9(e^2 + l)$ is equal to:
A.14$14$
B.15$15$
C.16$16$
D.12$12$
Solution
### Related Formula
For a standard hyperbola:
Focus: (pm ae, 0)$(\pm ae, 0)$
Directrix: x = pm fracae$x = \pm \frac{a}{e}$
Eccentricity relation: (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$
Length of latus rectum: l = frac2b^2a$l = \frac{2b^2}{a}$
### Core Logic
Given ae = sqrt10$ae = \sqrt{10}$ and fracae = frac9sqrt10$\frac{a}{e} = \frac{9}{\sqrt{10}}$. Multiplying these gives a^2$a^2$, which determines both parameters.
### Step 1: Find a and e
a^2 = (ae) cdot left(fracaeright) = sqrt10 cdot frac9sqrt10 = 9 implies a = 3$$a^2 = (ae) \cdot \left(\frac{a}{e}\right) = \sqrt{10} \cdot \frac{9}{\sqrt{10}} = 9 \implies a = 3$$
Substitute a = 3$a = 3$ into ae = sqrt10$ae = \sqrt{10}$:
e = fracsqrt103 implies e^2 = frac109$$e = \frac{\sqrt{10}}{3} \implies e^2 = \frac{10}{9}$$
### Step 2: Find b and l
Using (ae)^2 = a^2 + b^2$(ae)^2 = a^2 + b^2$:
10 = 9 + b^2 implies b^2 = 1$$10 = 9 + b^2 \implies b^2 = 1$$
Then the length of latus rectum l$l$ is:
l = frac2b^2a = frac2(1)3 = frac23$$l = \frac{2b^2}{a} = \frac{2(1)}{3} = \frac{2}{3}$$
### Step 3: Evaluate Final Expression
Calculate 9(e^2 + l)$9(e^2 + l)$:
9left(frac109 + frac23right) = 10 + 6 = 16$$9\left(\frac{10}{9} + \frac{2}{3}\right) = 10 + 6 = 16$$
### Pattern Recognition
Multiplying focus location by directrix location immediately eliminates e$e$, giving a^2$a^2$ directly. Once a^2$a^2$ is known, b^2$b^2$ follow seamlessly via (ae)^2 = a^2+b^2$(ae)^2 = a^2+b^2$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
More Conic Sections Questions — jee_main_2026_21_jan_morning
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