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 5
Q68jee_main_2025_03_april_morningCommon Tangents and Shortest Distance
The radius of the smallest circle which touches the parabolas y = x^2 + 2$y = x^{2} + 2$ and x = y^2 + 2$x = y^{2} + 2$ is[cite: 673]:
A.frac7sqrt22$\frac{7\sqrt{2}}{2}$
B.frac7sqrt216$\frac{7\sqrt{2}}{16}$
C.frac7sqrt24$\frac{7\sqrt{2}}{4}$
D.frac7sqrt28$\frac{7\sqrt{2}}{8}$
Solution
### Related Formula
Shortest distance between symmetric profiles: The minimal spacing normal line runs completely perpendicular to the mutual line of symmetry y=x$y=x$.
Common Tangents and Shortest Distance diagram for Q68 - JEE Main 2025 Morning
### Core Logic
The given curve equations reflect symmetry across line y=x$y=x$[cite: 1382, 1383]. The tangent slope at the closest matching locations must run parallel to this mirror path [cite: 1403]:
fracmathrmdymathrmdx = 1$$\frac{\mathrm{d}y}{\mathrm{d}x} = 1$$ [cite: 1403]
Differentiate curve equation y = x^2 + 2$y = x^2 + 2$ [cite: 1404]:
fracmathrmdymathrmdx = 2x = 1 implies x = frac12$$\frac{\mathrm{d}y}{\mathrm{d}x} = 2x = 1 \implies x = \frac{1}{2}$$ [cite: 1405, 1406]
Substitute back to get y-coordinate [cite: 1406]:
y = left(frac12right)^2 + 2 = frac94 implies Bleft(frac12, frac94right)$$y = \left(\frac{1}{2}\right)^2 + 2 = \frac{9}{4} \implies B\left(\frac{1}{2}, \frac{9}{4}\right)$$ [cite: 1406, 1407]
By mirror symmetry, the corresponding point on the other parabola is [cite: 1407]:
Aleft(frac94, frac12right)$$A\left(\frac{9}{4}, \frac{1}{2}\right)$$ [cite: 1407]
### Step 1: Calculating distance and circle radius
Evaluate chord distance AB$AB$ using standard metrics [cite: 1407]:
AB = sqrtleft(frac94 - frac12right)^2 + left(frac12 - frac94right)^2 = sqrt2 cdot left(frac74right)^2 = frac7sqrt24$$AB = \sqrt{\left(\frac{9}{4} - \frac{1}{2}\right)^2 + \left(\frac{1}{2} - \frac{9}{4}\right)^2} = \sqrt{2 \cdot \left(\frac{7}{4}\right)^2} = \frac{7\sqrt{2}}{4}$$ [cite: 1407, 1408]
The diameter of the smallest circle spanning between these touching curves equals distance AB$AB$ [cite: 1408].
textRadius = fracAB2 = frac7sqrt28$$\text{Radius} = \frac{AB}{2} = \frac{7\sqrt{2}}{8}$$ [cite: 1408]
### Pattern Recognition
Mutually inverse conic curves track symmetric footprints. Their closest distance segments always align perfectly perpendicular to the main baseline axis line y=x$y=x$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Conic Sections (Parabola)
Let the product of the focal distances of the point P(4, 2sqrt3)$P(4, 2\sqrt{3})$ on the hyperbola H : fracx^2a^2 - fracy^2b^2 = 1$H : \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be 32[cite: 690, 691]. Let the length of the conjugate axis of H$H$ be p$p$ and the length of its latus rectum be q$q$[cite: 692]. Then p^2 + q^2$p^2 + q^2$ is equal to[cite: 693]:
The axis of a parabola is the line y = x$y = x$ and its vertex and focus are in the first quadrant at distances sqrt2$\sqrt{2}$ and 2sqrt2$2\sqrt{2}$ units from the origin, respectively. If the point (1, k)$(1, k)$ lies on the parabola, then a possible value of k$k$ is:
A.4$4$
B.9$9$
C.3$3$
D.8$8$
Solution
### Related Formula
For any point P$P$ on a parabola, its distance to the focus S$S$ equals its perpendicular distance to the directrix line M$M$:
PS = PM$PS = PM$
### Core Logic
The axis line is y = x$y = x$. The vertex lies along this line at a distance of sqrt2$\sqrt{2}$ from the origin. Since it's in the first quadrant, its coordinates are (1,1)$(1,1)$.
The focus also lies along y=x$y=x$ at a distance of 2sqrt2$2\sqrt{2}$ from the origin, which gives coordinates (2,2)$(2,2)$.
### Step 1: Finding the Equation of the Directrix
The distance from the vertex to the focus is a = sqrt(2-1)^2 + (2-1)^2 = sqrt2$a = \sqrt{(2-1)^2 + (2-1)^2} = \sqrt{2}$.
The directrix is perpendicular to the axis line y = x$y = x$ (slope = 1$1$), so the slope of the directrix is -1$-1$.
The directrix is located at a distance a = sqrt2$a = \sqrt{2}$ behind the vertex, which brings it exactly to the origin (0,0)$(0,0)$.
Therefore, the equation of the directrix line is:
y - 0 = -1(x - 0) implies x + y = 0$$y - 0 = -1(x - 0) \implies x + y = 0$$Parabola diagram for Q57 - JEE Main 2025 Evening
### Step 2: Utilizing the Focus-Directrix Property
Let the point P(1,k)$P(1,k)$ lie on the parabola. Applying PS = PM$PS = PM$:
sqrt(1 - 2)^2 + (k - 2)^2 = frac|1 + k|sqrt1^2 + 1^2$$\sqrt{(1 - 2)^2 + (k - 2)^2} = \frac{|1 + k|}{\sqrt{1^2 + 1^2}}$$
Squaring both sides:
1 + (k - 2)^2 = frac(1 + k)^22$$1 + (k - 2)^2 = \frac{(1 + k)^2}{2}$$2big(1 + k^2 - 4k + 4big) = 1 + k^2 + 2k$$2\big(1 + k^2 - 4k + 4\big) = 1 + k^2 + 2k$$2k^2 - 8k + 10 = k^2 + 2k + 1$$2k^2 - 8k + 10 = k^2 + 2k + 1$$k^2 - 10k + 9 = 0$$k^2 - 10k + 9 = 0$$
Factoring the quadratic equations:
(k - 1)(k - 9) = 0 implies k = 1 text or k = 9$$(k - 1)(k - 9) = 0 \implies k = 1 \text{ or } k = 9$$
### Pattern Recognition
When a vertex and focus both sit perfectly on a symmetric line like y=x$y=x$, notice that the foot of the directrix often lands on a clean coordinate intersection (like the origin here), heavily simplifying geometric distance steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q60jee_main_2025_04_april_eveningHyperbola
Let the sum of the focal distances of the point mathrmP(4,3)$\mathrm{P}(4,3)$ on the hyperbola mathrmH:fracmathrmx^2mathrma^2 -fracmathrmy^2mathrmb^2 = 1$\mathrm{H}:\frac{\mathrm{x}^2}{\mathrm{a}^2} -\frac{\mathrm{y}^2}{\mathrm{b}^2} = 1$ be 8sqrtfrac53$8\sqrt{\frac{5}{3}}$. If for mathrmH$\mathrm{H}$, the length of the latus rectum is l$l$ and the product of the focal distances of the point mathrmP$\mathrm{P}$ is mathfrakm$\mathfrak{m}$, then 9l^2 + 6mathrmm$9l^{2} + 6\mathrm{m}$ is equal to:
A.184$184$
B.186$186$
C.185$185$
D.187$187$
Solution
### Related Formula
For a point P(x_1, y_1)$P(x_1, y_1)$ on a hyperbola branch, the focal distances are ex_1 + a$ex_1 + a$ and ex_1 - a$ex_1 - a$. Their sum is 2ex_1$2ex_1$, and their product is e^2x_1^2 - a^2$e^2x_1^2 - a^2$.
### Core Logic
Given the point P(4,3)$P(4,3)$, the x-coordinate is x_1 = 4$x_1 = 4$. The sum of focal distances is:
2ex_1 = 8sqrtfrac53 implies 2e(4) = 8sqrtfrac53 implies e = sqrtfrac53$$2ex_1 = 8\sqrt{\frac{5}{3}} \implies 2e(4) = 8\sqrt{\frac{5}{3}} \implies e = \sqrt{\frac{5}{3}}$$
Using the eccentricity relation b^2 = a^2(e^2 - 1)$b^2 = a^2(e^2 - 1)$:
b^2 = a^2left(frac53 - 1right) = frac23a^2$$b^2 = a^2\left(\frac{5}{3} - 1\right) = \frac{2}{3}a^2$$
### Step 1: Finding the Ellipse Parameters
Since P(4,3)$P(4,3)$ lies on the hyperbola fracx^2a^2 - fracy^2b^2 = 1$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:
frac16a^2 - frac9frac23a^2 = 1 implies frac16a^2 - frac272a^2 = 1$$\frac{16}{a^2} - \frac{9}{\frac{2}{3}a^2} = 1 \implies \frac{16}{a^2} - \frac{27}{2a^2} = 1$$frac32 - 272a^2 = 1 implies frac52a^2 = 1 implies a^2 = frac52$$\frac{32 - 27}{2a^2} = 1 \implies \frac{5}{2a^2} = 1 \implies a^2 = \frac{5}{2}$$
Now calculate b^2$b^2$:
b^2 = frac23left(frac52right) = frac53$$b^2 = \frac{2}{3}\left(\frac{5}{2}\right) = \frac{5}{3}$$
### Step 2: Calculating l^2 and m
The length of the latus rectum l$l$ is given by l = frac2b^2a$l = \frac{2b^2}{a}$:
l^2 = frac4b^4a^2 = frac4left(frac259right)frac52 = frac1009 times frac25 = frac409 implies 9l^2 = 40$$l^2 = \frac{4b^4}{a^2} = \frac{4\left(\frac{25}{9}\right)}{\frac{5}{2}} = \frac{100}{9} \times \frac{2}{5} = \frac{40}{9} \implies 9l^2 = 40$$
The product of focal distances m$m$ is:
m = e^2x_1^2 - a^2 = left(frac53right)(16) - frac52 = frac803 - frac52 = frac160 - 156 = frac1456$$m = e^2x_1^2 - a^2 = \left(\frac{5}{3}\right)(16) - \frac{5}{2} = \frac{80}{3} - \frac{5}{2} = \frac{160 - 15}{6} = \frac{145}{6}$$6m = 145$6m = 145$
### Step 3: Final Computation
Evaluating the targeted expression:
9l^2 + 6m = 40 + 145 = 185$$9l^2 + 6m = 40 + 145 = 185$$
### Pattern Recognition
Using focal property formulas directly (2ex_1$2ex_1$ for sum and e^2x_1^2 - a^2$e^2x_1^2 - a^2$ for product) avoids the lengthy process of finding focus coordinate values and executing distance formulas explicitly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q64jee_main_2025_04_april_eveningEllipse
Let for two distinct values of p$p$ the lines y = x + p$y = x + p$ touch the ellipse E: fracx^24^2 + fracy^23^2 = 1$E: \frac{x^2}{4^2} + \frac{y^2}{3^2} = 1$ at the points A and B. Let the line y = x$y = x$ intersect E at the points C and D. Then the area of the quadrilateral ABCD is equal to
A.36$36$
B.24$24$
C.48$48$
D.20$20$
Solution
### Related Formula
The condition for a line y = mx + p$y = mx + p$ to be tangent to an ellipse fracx^2a^2 + fracy^2b^2 = 1$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:
p^2 = a^2m^2 + b^2$$p^2 = a^2m^2 + b^2$$
The coordinate of the point of contact is given by left(-fraca^2mp, fracb^2pright)$\left(-\frac{a^2m}{p}, \frac{b^2}{p}\right)$.
### Core Logic
Given the ellipse parameter values a^2 = 16$a^2 = 16$ and b^2 = 9$b^2 = 9$, and tangent line slope m = 1$m = 1$:
p^2 = 16(1)^2 + 9 = 25 implies p = pm 5$$p^2 = 16(1)^2 + 9 = 25 \implies p = \pm 5$$
Thus, the two values of p$p$ are 5$5$ and -5$-5$. The points of contact A$A$ and B$B$ are:
- For p = 5$p = 5$: A = left(-frac16(1)5, frac95right) = left(-frac165, frac95right)$A = \left(-\frac{16(1)}{5}, \frac{9}{5}\right) = \left(-\frac{16}{5}, \frac{9}{5}\right)$
- For p = -5$p = -5$: B = left(-frac16(1)-5, frac9-5right) = left(frac165, -frac95right)$B = \left(-\frac{16(1)}{-5}, \frac{9}{-5}\right) = \left(\frac{16}{5}, -\frac{9}{5}\right)$
### Step 1: Intersecting line with Ellipse
The line y = x$y = x$ intersects the ellipse fracx^216 + fracy^29 = 1$\frac{x^2}{16} + \frac{y^2}{9} = 1$:
fracx^216 + fracx^29 = 1 implies frac25x^2144 = 1 implies x^2 = frac14425 implies x = pm frac125$$\frac{x^2}{16} + \frac{x^2}{9} = 1 \implies \frac{25x^2}{144} = 1 \implies x^2 = \frac{144}{25} \implies x = \pm \frac{12}{5}$$
Since y = x$y = x$, the intersection points C$C$ and D$D$ are:
C = left(-frac125, -frac125right) quad textand quad D = left(frac125, frac125right)$$C = \left(-\frac{12}{5}, -\frac{12}{5}\right) \quad \text{and} \quad D = \left(\frac{12}{5}, \frac{12}{5}\right)$$
### Step 2: Calculating Quadrilateral Area
The area of quadrilateral ABCD$ABCD$ with vertices mapped symmetrically can be computed using the standard coordinate determinant matrix layout formula:
textArea = frac12 beginvmatrix x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 endvmatrix + dots = 24$$\text{Area} = \frac{1}{2} \begin{vmatrix} x_A & y_A & 1 \\ x_B & y_B & 1 \\ x_C & y_C & 1 \end{vmatrix} + \dots = 24$$
### Pattern Recognition
Notice that the tangent lines are parallel and symmetric (p = pm 5$p = \pm 5$), and the intersecting line passes through the origin. This symmetry creates a geometric parallelogram, simplifying your area calculation by doubling the area of triangle ABD$ABD$.
### 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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