If the equation of the parabola with vertex V((3/2),3) and the directrix x+2y=0 is alpha x2+beta y2-gamma…

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Q11 jee_main_2026_21_jan_morning Coinciding Foci of Ellipse and Hyperbola

Let the foci of hyperbola coincide with the foci of the ellipse

(x²)/(36) +(y²)/(16) = 1

. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is:

A. 12
B. 16
C. $96\sqrt(5)$ $\frac{96}{\sqrt{5}}$
D. $24\sqrt(5)$ $24\sqrt{5}$

Solution

Related Formula

Eccentricity of ellipse e₁ = \sqrt(1 - (b²)/(a²))

Foci = (\pm ae₁, 0)

Length of Latus Rectum of hyperbola = 2bhyp²ahyp

Core Logic

For the given ellipse $(x²)/(36) + (y²)/(16) = 1$ $\frac{x^2}{36} + \frac{y^2}{16} = 1$: $a² = 36 \Rightarrow a = 6$ $a^2 = 36 \Rightarrow a = 6$ $b² = 16$ $b^2 = 16$

e₁ = \sqrt(1 - (16)/(36)) = \sqrt(1 - (4)/(9)) = \sqrt(5)3

Foci of the ellipse are at $(\pm ae₁, 0) = (\pm 6 \cdot \sqrt(5)3, 0) = (\pm 2\sqrt(5), 0)$ .

Step 1: Establish Hyperbola Parameters

Let the hyperbola be $(x²)/(p²) - (y²)/(q²) = 1$ $\frac{x^2}{p^2} - \frac{y^2}{q^2} = 1$. Its foci coincide with the ellipse, so the foci of hyperbola are also $(\pm 2\sqrt(5), 0)$ $(\pm 2\sqrt{5}, 0)$. Let $e$ $e$ be the eccentricity of the hyperbola. We are given $e = 5$ $e = 5$. Focus of hyperbola is $pe = 2\sqrt(5)$ $pe = 2\sqrt{5}$.

p(5) = 2\sqrt(5) \Rightarrow p = 2\sqrt(5)5 = 2\sqrt(5)

Step 2: Find the Conjugate Axis (q)

For the hyperbola:

e² = 1 + (q²)/(p²)

25 = 1 + q²( 2\sqrt(5))²

24 = (q²)/(4/5) \Rightarrow 24 = (5q²)/(4)

5q² = 96 \Rightarrow q² = (96)/(5)

Step 3: Calculate Latus Rectum

Length of Latus Rectum = $(2q²)/(p)$ $\frac{2q^2}{p}$

= 2 ((96)/(5)) 2\sqrt(5) = (96)/(5) \times \sqrt(5) = 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.

Chapter Mix

Class 11 Maths: Conic Sections

Q15 jee_main_2026_21_jan_morning Locus of Intersection of Tangents

Let PQ and MN be two straight lines touching the circle

x² + y² - 4x - 6y - 3 = 0

at the points A and B respectively. Let O be the centre of the circle and

∠ AOB = π/3

. Then the locus of the point of intersection of the lines PQ and MN is:

A. $3(x² + y²) - 18x - 12y + 25 = 0$ $3(x^{2} + y^{2}) - 18x - 12y + 25 = 0$
B. $x² + y² - 12x - 18y - 25 = 0$ $x^{2} + y^{2} - 12x - 18y - 25 = 0$
C. $x² + y² - 18x - 12y - 25 = 0$ $x^{2} + y^{2} - 18x - 12y - 25 = 0$
D. $3(x² + y²) - 12x - 18y - 25 = 0$ $3(x^{2} + y^{2}) - 12x - 18y - 25 = 0$

Solution

Related Formula

For external tangents from point $R$ $R$ forming angle $2θ$ $2\theta$ at the center, the distance $d$ $d$ from center to intersection point obeys $θ = (r)/(d)$ $\cos \theta = \frac{r}{d}$.

Core Logic

Given circle: $x² + y² - 4x - 6y - 3 = 0$ $x^2 + y^2 - 4x - 6y - 3 = 0$ Center $O = (2, 3)$ $O = (2, 3)$ Radius $r = \sqrt((-2)² + (-3)² - (-3)) = \sqrt(4 + 9 + 3) = \sqrt(16) = 4$ .

The tangents PQ and MN intersect at some point $R(h, k)$ $R(h, k)$. The radius vectors $OA$ $OA$ and $OB$ $OB$ subtend angle $∠ AOB = (π)/(3) = 60°$ at the center. The line joining $O$ $O$ and $R$ $R$ bisects the angle $∠ AOB$ $\angle AOB$. Thus, $∠ AOR = 30°$ $\angle AOR = 30^{\circ}$.

Step 1: Apply Trigonometric Relations

Circle tangents intersection locus diagram for Q15 - JEE Main 2026 Morning

In the right-angled triangle $Δ AOR$ $\Delta AOR$, $OA$ $OA$ is the radius ( $r = 4$ $r = 4$) and $OR$ $OR$ is the hypotenuse.

(30°) = (OA)/(OR) = (r)/(OR)

\sqrt(3)2 = (4)/(OR) \Rightarrow OR = 8\sqrt(3)

Step 2: Construct the Locus Equation

The distance squared between $O(2,3)$ $O(2,3)$ and $R(h,k)$ $R(h,k)$ is $OR²$ $OR^2$:

OR² = (h - 2)² + (k - 3)² = ( 8\sqrt(3))²

(h - 2)² + (k - 3)² = (64)/(3)

h² - 4h + 4 + k² - 6k + 9 = (64)/(3)

3(h² + k² - 4h - 6k + 13) = 64

3h² + 3k² - 12h - 18k + 39 - 64 = 0

3(h² + k²) - 12h - 18k - 25 = 0

Step 3: Generalize the Equation

Replace $(h, k)$ $(h, k)$ with $(x, y)$ $(x, y)$ for the general locus:

3(x² + y²) - 12x - 18y - 25 = 0

Pattern Recognition

The locus of the intersection of tangents enclosing a constant angle is simply a concentric circle. Its radius expands by $1/ (α/2)$ $1/\sin(\alpha/2)$ or $1/ (θ)$ $1/\cos(\theta)$ depending on whether the angle is measured at intersection or center.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

Q20 jee_main_2026_21_jan_morning Locus of Internal Section Point

Let O be the vertex of the parabola

x²=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

Section Formula: P = (m \cdot Q + n \cdot O)/(m + n)

Chord bisected at (x₁, y₁) : T = S₁

Core Logic

Given parabola $x² = 4y$ $x^2 = 4y$, its vertex $O = (0, 0)$ $O = (0, 0)$. A general point $Q$ $Q$ on $x² = 4y$ $x^2 = 4y$ is $(2t, t²)$ $(2t, t^2)$. Let $P(h, k)$ $P(h, k)$ divide $OQ$ $OQ$ in ratio $2:3$ $2:3$. By section formula:

h = (2(2t) + 3(0))/(5) = (4t)/(5)

k = (2(t²) + 3(0))/(5) = (2t²)/(5)

Locus of point dividing parabolic chord for Q20 - JEE Main 2026 Morning

Step 1: Finding the Locus C

From $h = (4t)/(5)$ $h = \frac{4t}{5}$, we get $t = (5h)/(4)$ $t = \frac{5h}{4}$. Substitute into $k$ $k$:

k = (2)/(5) ((5h)/(4))² = (2)/(5) \cdot (25h²)/(16) = (5h²)/(8)

8k = 5h² \Rightarrow 5x² = 8y

So the conic C is the parabola $5x² = 8y$ $5x^2 = 8y$.

Step 2: Chord bisected at a point

We need the equation of the chord of $C: 5x² - 8y = 0$ $C: 5x^2 - 8y = 0$ bisected at $(x₁, y₁) = (1, 2)$ $(x_1, y_1) = (1, 2)$. Use $T = S₁$ $T = S_1$. $T = 5xx₁ - 4(y + y₁) = 5x(1) - 4(y + 2) = 5x - 4y - 8$ $S₁ = 5(1)² - 8(2) = 5 - 16 = -11$ $S_1 = 5(1)^2 - 8(2) = 5 - 16 = -11$

Equating $T$ $T$ and $S₁$ $S_1$:

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 ( $T=S₁$ $T=S_1$) strictly applies algebraically.

Chapter Mix

Class 11 Maths: Parabola Class 11 Maths: Straight Lines

Q4 jee_main_2026_21_jan_evening Ellipse

In the line

α x + 4y = \sqrt(7)

, where

α in R

, touches the ellipse

3x² + 4y² = 1

at the point

P

in the first quadrant, then one of the focal distances of

P

is:

A. $1\sqrt(3) - 12\sqrt(11)$
B. $1\sqrt(3) + 12\sqrt(5)$
C. $1\sqrt(3) - 12\sqrt(5)$
D. $1\sqrt(3) + 12\sqrt(7)$

Solution

Related Formula

Condition of tangency for ellipse (x²)/(a²) + (y²)/(b²) = 1 is c² = a²m² + b²

Focal distance SP = a \pm ex

Eccentricity e = \sqrt(1 - (b²)/(a²))

Core Logic

Ellipse diagram for Q4 - JEE Main 2026 Evening

Identify the slope and intercepts of the line to find

α

. Use the point of contact formula to locate

P(x₁, y₁)

and apply focal distance definitions.

Step 1: Determine alpha

Rewrite the ellipse: $(x²)/(1/3) + (y²)/(1/4) = 1 a² = (1)/(3), b² = (1)/(4)$ . The line is $y = -(α)/(4)x + \sqrt(7)4$ . Using $c² = a²m² + b²$ $c^2 = a^2m^2 + b^2$:

( \sqrt(7)4)² = (1)/(3) (-(α)/(4))² + (1)/(4)

(7)/(16) = (α²)/(48) + (4)/(16) (3)/(16) = (α²)/(48) α² = 9 α = \pm 3

Since $P$ $P$ is in the first quadrant, coordinates $x, y$ $x, y$ are positive, so we use the tangent $3x + 4y - \sqrt(7) = 0$ $3x + 4y - \sqrt{7} = 0$.

Step 2: Find Point of Contact P

The tangent at $P(x₁, y₁)$ $P(x_1, y_1)$ is $3xx₁ + 4yy₁ = 1$ $3xx_1 + 4yy_1 = 1$. Comparing this with $3x + 4y = \sqrt(7)$ $3x + 4y = \sqrt{7}$ (divided by $\sqrt(7)$ $\sqrt{7}$ to match constant $1$ $1$): $3x\sqrt(7) + 4y\sqrt(7) = 1$ . Comparing coefficients:

3x₁ = 3\sqrt(7) x₁ = 1\sqrt(7)

4y₁ = 4\sqrt(7) y₁ = 1\sqrt(7)

So $P = ( 1\sqrt(7), 1\sqrt(7))$ .

Step 3: Calculate Focal Distance

Find eccentricity:

e = \sqrt(1 - (1/4)/(1/3)) = \sqrt(1 - (3)/(4)) = (1)/(2)

The focal distances are $a - ex$ $a - ex$ and $a + ex$ $a + ex$. Since $a² = 1/3 a = 1/\sqrt(3)$ $a^2 = 1/3 \implies a = 1/\sqrt{3}$.

SP = a - ex₁ = 1\sqrt(3) - (1)/(2)( 1\sqrt(7)) = 1\sqrt(3) - 12\sqrt(7)

S'P = a + ex₁ = 1\sqrt(3) + (1)/(2)( 1\sqrt(7)) = 1\sqrt(3) + 12\sqrt(7)

Matching with the options, the focal distance is $1\sqrt(3) + 12\sqrt(7)$ .

Pattern Recognition

For tangency $lx+my+n=0$ $lx+my+n=0$ to $x²/a² + y²/b² = 1$ $x^2/a^2 + y^2/b^2 = 1$, use $a² l² + b² m² = n²$ $a^2 l^2 + b^2 m^2 = n^2$. Points of contact can be quickly evaluated by comparing $T=0$ $T=0$ to the normalized tangent equation.

Chapter Mix

Class 11 Maths: Conic Sections

Q5 jee_main_2026_21_jan_evening Parabola

Let

y² = 12x

be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that

∠ OPA = 90°

. Then the locus of the centroid of such triangles OPA is:

A. $y² - 6x + 4 = 0$ $y^{2} - 6x + 4 = 0$
B. $y² - 9x + 6 = 0$ $y^{2} - 9x + 6 = 0$
C. $y² - 2x + 8 = 0$ $y^{2} - 2x + 8 = 0$
D. $y² - 4x + 8 = 0$ $y^{2} - 4x + 8 = 0$

Solution

Related Formula

Centroid G(x,y) = ( (x₁ + x₂ + x₃)/(3), (y₁ + y₂ + y₃)/(3) )

Product of slopes for perpendicular lines m₁ m₂ = -1

Core Logic

Parabola geometry diagram for Q5 - JEE Main 2026 Evening

Define coordinates:

O(0,0)

,

P(3t², 6t)

(since

y² = 12x 4a = 12 a=3

). Determine the slope of

OP

and use perpendicularity to find the equation of

PA

and locate point

A

on the x-axis.

Step 1: Locate A via Perpendicularity

Slope of $OP$ $OP$ is $mOP = (6t - 0)/(3t² - 0) = (2)/(t)$ . Since $∠ OPA = 90^°$ $\angle OPA = 90^\circ$, slope of $AP$ $AP$ is $mAP = -(t)/(2)$ $m_{AP} = -\frac{t}{2}$. Equation of $AP$ $AP$:

y - 6t = -(t)/(2)(x - 3t²)

To find $A$ $A$ on the x-axis, put $y = 0$ $y = 0$:

-6t = -(t)/(2)(x - 3t²) 12 = x - 3t² x = 12 + 3t²

So, $A$ $A$ is $(12 + 3t², 0)$ $(12 + 3t^2, 0)$.

Step 2: Locus of the Centroid

Let the centroid of $OPA$ $\triangle OPA$ be $G(h, k)$ $G(h, k)$.

h = (0 + 3t² + (12 + 3t²))/(3) = (6t² + 12)/(3) = 2t² + 4

k = (0 + 6t + 0)/(3) = 2t

From $k = 2t t = (k)/(2)$ $k = 2t \implies t = \frac{k}{2}$. Substitute $t$ $t$ into the equation for $h$ $h$:

h = 2((k)/(2))² + 4 = (k²)/(2) + 4

2h = k² + 8 k² = 2h - 8

Replacing $(h, k)$ $(h, k)$ with $(x, y)$ $(x, y)$, the locus is $y² = 2x - 8 y² - 2x + 8 = 0$ .

Pattern Recognition

For any right-angled configuration involving the origin and axis on a parabola, parametric geometry simplifies equations significantly. Find coordinates $O, P, A$ $O, P, A$, set up the centroid algebraic relations, and eliminate parameter $t$ $t$.

Chapter Mix

Class 11 Maths: Conic Sections Class 11 Maths: Straight Lines

If the equation of the parabola with vertex V((3)/(2),3)$V\left(\frac{3}{2},3\right)$ and the directrix x+2y=0$x+2y=0$ is α x²+β y²-γ xy-30x-60y+225=0,$\alpha x^{2}+\beta y^{2}-\gamma xy-30x-60y+225=0,$ then α+β+γ$\alpha+\beta+\gamma$ is equal to:

Solution & Explanation

Related Formula

Step 1: Determine the Focus coordinates

Step 2: Derive the Parabola Locus Equation

Step 3: Coefficient Extraction

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions

Solution

Related Formula

Core Logic

Step 1: Establish Hyperbola Parameters

Step 2: Find the Conjugate Axis (q)

Step 3: Calculate Latus Rectum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Apply Trigonometric Relations

Step 2: Construct the Locus Equation

Step 3: Generalize the Equation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding the Locus C

Step 2: Chord bisected at a point

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine alpha

Step 2: Find Point of Contact P

Step 3: Calculate Focal Distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Locate A via Perpendicularity

Step 2: Locus of the Centroid

Pattern Recognition

Chapter Mix

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If the equation of the parabola with vertex $V((3)/(2),3)$ $V\left(\frac{3}{2},3\right)$ and the directrix $x+2y=0$ $x+2y=0$ is $α x²+β y²-γ xy-30x-60y+225=0,$ then $α+β+γ$ $\alpha+\beta+\gamma$ is equal to: