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 2
Qjee_main_2025_02_april_morningProperties of Ellipse
If S$S$ and S'$S'$ are the foci of the ellipse fracx^218 + fracy^29 = 1$\frac{x^2}{18} + \frac{y^2}{9} = 1$ and P$P$ be a point on the ellipse, then min(SP cdot S'P) + max(SP cdot S'P)$\min(SP \cdot S'P) + \max(SP \cdot S'P)$ is equal to:
A.3(1+sqrt2)$3(1+\sqrt{2})$
B.3(6+sqrt2)$3(6+\sqrt{2})$
C.9$9$
D.27$27$
Solution
### Related Formula
Focal distances of any point P(acostheta, bsintheta)$P(a\cos\theta, b\sin\theta)$ on an ellipse are given by:
SP = a - ex_P = a(1 - ecostheta)$$SP = a - ex_P = a(1 - e\cos\theta)$$S'P = a + ex_P = a(1 + ecostheta)$$S'P = a + ex_P = a(1 + e\cos\theta)$$
Product of focal distances:
SP cdot S'P = a^2(1 - e^2cos^2theta) = a^2 - e^2x_P^2$$SP \cdot S'P = a^2(1 - e^2\cos^2\theta) = a^2 - e^2x_P^2$$
### Core Logic
Compute the eccentricity e$e$, express the product SP cdot S'P$SP \cdot S'P$ in terms of cos^2theta$\cos^2\theta$, and analyze its bounds across the domain to find minimum and maximum limits. Properties of Ellipse diagram for Q66 - JEE Main 2025 Morning
### Step 1: Determine Ellipse Parameters
Given a^2 = 18$a^2 = 18$ and b^2 = 9$b^2 = 9$.
b^2 = a^2(1 - e^2) implies 9 = 18(1 - e^2) implies 1 - e^2 = frac12 implies e = frac1sqrt10$$b^2 = a^2(1 - e^2) \implies 9 = 18(1 - e^2) \implies 1 - e^2 = \frac{1}{2} \implies e = \frac{1}{\sqrt{10}}$$
### Step 2: Express Focal Product
The parametric coordinates are P(3sqrt2costheta, 3sintheta)$P(3\sqrt{2}\cos\theta, 3\sin\theta)$.
SP cdot S'P = a^2 - (ae)^2cos^2theta$$SP \cdot S'P = a^2 - (ae)^2\cos^2\theta$$
Since a^2=18$a^2=18$ and (ae)^2 = a^2-b^2 = 18-9 = 9$(ae)^2 = a^2-b^2 = 18-9 = 9$:
SP cdot S'P = 18 - 9cos^2theta$$SP \cdot S'P = 18 - 9\cos^2\theta$$
### Step 3: Evaluate Extrema and Sum
Since 0 le cos^2theta le 1$0 \le \cos^2\theta \le 1$:
* Maximum value occurs when cos^2theta = 0 implies max = 18$\cos^2\theta = 0 \implies \max = 18$.
* Minimum value occurs when cos^2theta = 1 implies min = 18 - 9 = 9$\cos^2\theta = 1 \implies \min = 18 - 9 = 9$.
textSum = min + max = 9 + 18 = 27$$\text{Sum} = \min + \max = 9 + 18 = 27$$
### Pattern Recognition
The product of focal distances can also be written directly as b^2$b^2$ at the minor axis vertices (max) and a^2(1-e^2)$a^2(1-e^2)$ varying down to a^2-c^2$a^2-c^2$. Summing them up yields b^2 + a^2 = 9 + 18 = 27$b^2 + a^2 = 9 + 18 = 27$ instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Qjee_main_2025_02_april_morningProperties of Parabola
Let the focal chord PQ$PQ$ of the parabola y^2 = 4x$y^2 = 4x$ make an angle of 60^circ$60^\circ$ with the positive x$x$-axis, where P$P$ lies in the first quadrant. If the circle, whose one diameter is PS$PS$, S$S$ being the focus of the parabola, touches the y$y$-axis at the point (0, a)$(0, a)$, then 5a^2$5a^2$ is equal to:
A.15$15$
B.25$25$
C.30$30$
D.20$20$
Solution
### Related Formula
For a standard parabola y^2 = 4ax$y^2 = 4ax$:
Focus: S(a, 0)$S(a, 0)$
Parametric coordinates: (at^2, 2at)$(at^2, 2at)$
Equation of a circle on diametric endpoints (x_1, y_1)$(x_1, y_1)$ and (x_2, y_2)$(x_2, y_2)$:
(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$
### Core Logic
Find the point P$P$ using the slope of the focal chord, write the equation of the circle with diameter PS$PS$, and find its y$y$-intercept. Properties of Parabola diagram for Q70 - JEE Main 2025 Morning
### Step 1: Determine P Coordinates
For y^2 = 4x$y^2 = 4x$, parameter a=1 implies S(1,0)$a=1 \implies S(1,0)$ and P(t^2, 2t)$P(t^2, 2t)$.
Slope of focal chord PS$PS$:
tan 60^circ = frac2t - 0t^2 - 1 = sqrt3 implies 2t = sqrt3t^2 - sqrt3$$\tan 60^\circ = \frac{2t - 0}{t^2 - 1} = \sqrt{3} \implies 2t = \sqrt{3}t^2 - \sqrt{3}$$sqrt3t^2 - 2t - sqrt3 = 0 implies (sqrt3t + 1)(t - sqrt3) = 0$$\sqrt{3}t^2 - 2t - \sqrt{3} = 0 \implies (\sqrt{3}t + 1)(t - \sqrt{3}) = 0$$
Since P$P$ is in the first quadrant, t > 0 implies t = sqrt3$t > 0 \implies t = \sqrt{3}$.
Thus, P((sqrt3)^2, 2sqrt3) = P(3, 2sqrt3)$P((\sqrt{3})^2, 2\sqrt{3}) = P(3, 2\sqrt{3})$.
### Step 2: Construct the Diametric Circle Equation
Endpoints are S(1,0)$S(1,0)$ and P(3, 2sqrt3)$P(3, 2\sqrt{3})$:
(x - 1)(x - 3) + (y - 0)(y - 2sqrt3) = 0$$(x - 1)(x - 3) + (y - 0)(y - 2\sqrt{3}) = 0$$
### Step 3: Solve for y-intercept
The circle touches/intersects the y$y$-axis at x = 0$x = 0$:
(0 - 1)(0 - 3) + y(y - 2sqrt3) = 0 implies 3 + y^2 - 2sqrt3y = 0$$(0 - 1)(0 - 3) + y(y - 2\sqrt{3}) = 0 \implies 3 + y^2 - 2\sqrt{3}y = 0$$
This is a perfect square expression (y - sqrt3)^2 = 0 implies y = sqrt3$(y - \sqrt{3})^2 = 0 \implies y = \sqrt{3}$.
Thus, the intercept value is a = sqrt3$a = \sqrt{3}$.
### Step 4: Compute Final Target Value
5a^2 = 5(sqrt3)^2 = 15$$5a^2 = 5(\sqrt{3})^2 = 15$$
### Pattern Recognition
A circle whose diameter is a focal radius always touches the tangent at the vertex (y$y$-axis for a standard parabola). The coordinate of the contact point is simply given by a t = 1 cdot sqrt3 = sqrt3$a t = 1 \cdot \sqrt{3} = \sqrt{3}$, bypasses the full equation construction entirely.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q53jee_main_2025_03_april_eveningCircles
If the four distinct points (4, 6)$(4, 6)$, (-1, 5)$(-1, 5)$, (0, 0)$(0, 0)$ and (k, 3k)$(k, 3k)$ lie on a circle of radius r$r$, then 10k + r^2$10k + r^2$ is equal to
A.32$32$
B.33$33$
C.34$34$
D.35$35$
Solution
### Related Formula
The general equation of a circle is:
x^2 + y^2 + 2gx + 2fy + c = 0$$x^2 + y^2 + 2gx + 2fy + c = 0$$
Radius of the circle:
r = sqrtg^2 + f^2 - c$$r = \sqrt{g^2 + f^2 - c}$$
If a set of points lies on this circle, their coordinates must satisfy the equation.
### Core Logic
Since (0,0)$(0,0)$ lies on the circle:
0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 implies c = 0$$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0 \implies c = 0$$
Thus, the equation simplifies to:
x^2 + y^2 + 2gx + 2fy = 0$$x^2 + y^2 + 2gx + 2fy = 0$$
### Step 1: Finding g$g$, f$f$ and r^2$r^2$
Substitute (4,6)$(4,6)$:
16 + 36 + 8g + 12f = 0 implies 2g + 3f = -13 quad text--- (1)$$16 + 36 + 8g + 12f = 0 \implies 2g + 3f = -13 \quad \text{--- (1)}$$
Substitute (-1,5)$(-1,5)$:
1 + 25 - 2g + 10f = 0 implies -g + 5f = -13 implies g = 5f + 13 quad text--- (2)$$1 + 25 - 2g + 10f = 0 \implies -g + 5f = -13 \implies g = 5f + 13 \quad \text{--- (2)}$$
Substituting g$g$ from (2) into (1):
2(5f + 13) + 3f = -13$$2(5f + 13) + 3f = -13$$13f + 26 = -13 implies f = -3$$13f + 26 = -13 \implies f = -3$$g = 5(-3) + 13 = -2$$g = 5(-3) + 13 = -2$$
The circle equation is:
x^2 + y^2 - 4x - 6y = 0$$x^2 + y^2 - 4x - 6y = 0$$
Calculating radius squared r^2$r^2$:
r^2 = g^2 + f^2 - c = (-2)^2 + (-3)^2 - 0 = 13$$r^2 = g^2 + f^2 - c = (-2)^2 + (-3)^2 - 0 = 13$$Circle diagram for Q53 - JEE Main 2025 Evening Shift
### Step 2: Solving for k$k$
The point (k, 3k)$(k, 3k)$ lies on this circle:
k^2 + (3k)^2 - 4k - 6(3k) = 0$$k^2 + (3k)^2 - 4k - 6(3k) = 0$$10k^2 - 22k = 0 implies k(10k - 22) = 0$$10k^2 - 22k = 0 \implies k(10k - 22) = 0$$
Since the points must be distinct and k=0$k=0$ gives (0,0)$(0,0)$ which is already a given point, we must have:
10k = 22 implies k = frac115$$10k = 22 \implies k = \frac{11}{5}$$
Now, calculate 10k + r^2$10k + r^2$:
10k + r^2 = 10left(frac115right) + 13 = 22 + 13 = 35$$10k + r^2 = 10\left(\frac{11}{5}\right) + 13 = 22 + 13 = 35$$
### Pattern Recognition
Notice that the slope of the line joining origin (0,0)$(0,0)$ to the general point is y = 3x$y = 3x$. For three given coordinates, if origin is one of them, the circle equation lacks the constant c$c$. It is always faster to first solve for parameters g, f$g, f$ and then check geometry.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 10 Mathematics: Coordinate Geometry
Q67jee_main_2025_03_april_eveningEllipse
Let C$C$ be the circle of minimum area enclosing the ellipse E: fracx^2a^2 + fracy^2b^2 = 1$E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with eccentricity frac12$\frac{1}{2}$ and foci (pm 2, 0)$(pm 2, 0)$. Let PQR$PQR$ be a variable triangle, whose vertex P$P$ is on the circle C$C$ and the side QR$QR$ of length 8$8$ is parallel to the major axis of E$E$ and contains the point of intersection of E$E$ with the negative y$y$-axis. Then the maximum area of the triangle PQR$PQR$ is:
A.6(3 + sqrt2)$6(3 + \sqrt{2})$
B.8(3 + sqrt2)$8(3 + \sqrt{2})$
C.6(2 + sqrt3)$6(2 + \sqrt{3})$
D.8(2 + sqrt3)$8(2 + \sqrt{3})$
Solution
### Related Formula
For an ellipse E$E$:
- Foci: (pm ae, 0)$(\pm ae, 0)$
- Eccentricity: b^2 = a^2(1 - e^2)$b^2 = a^2(1 - e^2)$
- The circle of minimum area enclosing a centered ellipse has diameter equal to the major axis of the ellipse (R = a$R = a$).
- Area of triangle: textArea = frac12 cdot textbase cdot textheight$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}$
### Core Logic
Let's first find coordinates a$a$ and b$b$:
- ae = 2$ae = 2$
- e = frac12 implies aleft(frac12right) = 2 implies a = 4$e = \frac{1}{2} \implies a\left(\frac{1}{2}\right) = 2 \implies a = 4$
- b^2 = a^2(1 - e^2) = 16left(1 - frac14right) = 12 implies b = 2sqrt3$b^2 = a^2(1 - e^2) = 16\left(1 - \frac{1}{4}\right) = 12 \implies b = 2\sqrt{3}$
### Step 1: Setting Circle and Triangle geometry
The enclosing circle C$C$ has radius R = a = 4$R = a = 4$, centered at (0,0)$(0,0)$. Thus, its equation is:
x^2 + y^2 = 16 implies P = (4costheta, 4sintheta)$$x^2 + y^2 = 16 \implies P = (4\cos\theta, 4\sin\theta)$$
The intersection of the ellipse with the negative y$y$-axis is (0, -b) = (0, -2sqrt3)$(0, -b) = (0, -2\sqrt{3})$.
Since side QR$QR$ (length = 8$= 8$) is parallel to the major axis (x$x$-axis) and contains (0, -2sqrt3)$(0, -2\sqrt{3})$, the equation of the line containing QR$QR$ is:
y = -2sqrt3$y = -2\sqrt{3}$Enclosing circle diagram for Q67 - JEE Main 2025 Evening Shift
### Step 2: Maximizing Area of Delta PQR$\Delta PQR$
The perpendicular height of vertex P(4costheta, 4sintheta)$P(4\cos\theta, 4\sin\theta)$ from the base line y = -2sqrt3$y = -2\sqrt{3}$ is:
H = 4sintheta - (-2sqrt3) = 4sintheta + 2sqrt3$$H = 4\sin\theta - (-2\sqrt{3}) = 4\sin\theta + 2\sqrt{3}$$
To maximize the area, we maximize height H$H$ by choosing sintheta = 1$\sin\theta = 1$:
H_max = 4 + 2sqrt3$$H_{max} = 4 + 2\sqrt{3}$$textMaximum Area = frac12 cdot textbase QR cdot H_max$$\text{Maximum Area} = \frac{1}{2} \cdot \text{base } QR \cdot H_{max}$$textMaximum Area = frac12 cdot 8 cdot (4 + 2sqrt3) = 4(4 + 2sqrt3) = 8(2 + sqrt3)$$\text{Maximum Area} = \frac{1}{2} \cdot 8 \cdot (4 + 2\sqrt{3}) = 4(4 + 2\sqrt{3}) = 8(2 + \sqrt{3})$$
### Pattern Recognition
The enclosing circle with minimum area is called the auxiliary circle. Its radius is equal to the semi-major axis a$a$. Max height of a triangle with a base fixed at line y=-k$y=-k$ and vertex on the circle is R + k$R + k$. This directly gives textArea = frac12 cdot textbase cdot (a+b)$\text{Area} = \frac{1}{2} \cdot \text{base} \cdot (a+b)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Q68jee_main_2025_03_april_eveningParabola
The shortest distance between the curves y^2 = 8x$y^2 = 8x$ and x^2 + y^2 + 12y + 35 = 0$x^2 + y^2 + 12y + 35 = 0$ is :
A.2sqrt3 - 1$2\sqrt{3} - 1$
B.sqrt2$\sqrt{2}$
C.3sqrt2 - 1$3\sqrt{2} - 1$
D.2sqrt2 - 1$2\sqrt{2} - 1$
Solution
### Related Formula
For a circle x^2 + (y-k)^2 = R^2$x^2 + (y-k)^2 = R^2$ and any smooth curve, the shortest distance lies along the normal to the curve passing through the center of the circle C(h,k)$C(h,k)$:
textShortest Distance = textDistance(P, C) - R$$\text{Shortest Distance} = \text{Distance}(P, C) - R$$
where P$P$ is the point of normal intersection on the curve.
### Core Logic
Let's first identify the circle parameters:
x^2 + y^2 + 12y + 35 = 0 implies x^2 + (y+6)^2 = 36 - 35 = 1$$x^2 + y^2 + 12y + 35 = 0 \implies x^2 + (y+6)^2 = 36 - 35 = 1$$
Center C = (0, -6)$C = (0, -6)$ and radius R = 1$R = 1$.
The first curve is the parabola y^2 = 8x$y^2 = 8x$, where a = 2$a = 2$.
Normal equation of y^2 = 4ax$y^2 = 4ax$ in slope form:
y = mx - 2am - am^3$$y = mx - 2am - am^3$$
Substituting a=2$a=2$:
y = mx - 4m - 2m^3$$y = mx - 4m - 2m^3$$
### Step 1: Find normal passing through circle center
Normal passes through C(0, -6)$C(0, -6)$:
-6 = m(0) - 4m - 2m^3$$-6 = m(0) - 4m - 2m^3$$2m^3 + 4m - 6 = 0 implies m^3 + 2m - 3 = 0$$2m^3 + 4m - 6 = 0 \implies m^3 + 2m - 3 = 0$$
By inspection, m=1$m=1$ is a real solution:
(m-1)(m^2 + m + 3) = 0$$(m-1)(m^2 + m + 3) = 0$$
Since m^2 + m + 3 = 0$m^2 + m + 3 = 0$ has complex roots, the unique real normal slope is m=1$m=1$.
Shortest distance diagram for Q68 - JEE Main 2025 Evening Shift
### Step 2: Point calculation and shortest distance
For m=1$m=1$ and a=2$a=2$, normal intersection point P(am^2, -2am)$P(am^2, -2am)$ is:
P = (2(1)^2, -2(2)(1)) = (2, -4)$$P = (2(1)^2, -2(2)(1)) = (2, -4)$$
Distance from P(2,-4)$P(2,-4)$ to center C(0,-6)$C(0,-6)$:
PC = sqrt(2-0)^2 + (-4 - (-6))^2 = sqrt4 + 4 = 2sqrt2$$PC = \sqrt{(2-0)^2 + (-4 - (-6))^2} = \sqrt{4 + 4} = 2\sqrt{2}$$
Shortest distance:
textSD = PC - R = 2sqrt2 - 1$$\text{SD} = PC - R = 2\sqrt{2} - 1$$
### Pattern Recognition
The shortest distance between a parabola and a circle is always along the common normal of the parabola passing through the circle's center. Finding the normal in slope form and solving for m$m$ avoids complex calculus.
### 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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