A line passing through the point P(sqrt(5), sqrt(5)) intersects the ellipse (x2/36) + (y2/25) = 1 at A and B [cite…

Year	2026	2025	2024	Total
Questions	29	44	21	94

Q53 jee_main_2025_03_april_evening Circles

If the four distinct points

(4, 6)

,

(-1, 5)

,

(0, 0)

and

(k, 3k)

lie on a circle of radius

r

, then

10k + r²

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:

x² + y² + 2gx + 2fy + c = 0

Radius of the circle:

r = \sqrt(g² + f² - 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:

0² + 0² + 2g(0) + 2f(0) + c = 0 c = 0

Thus, the equation simplifies to:

x² + y² + 2gx + 2fy = 0

Step 1: Finding $g$ $g$, $f$ $f$ and $r²$ $r^2$

Substitute $(4,6)$ $(4,6)$:

16 + 36 + 8g + 12f = 0 2g + 3f = -13 --- (1)

Substitute $(-1,5)$ $(-1,5)$:

1 + 25 - 2g + 10f = 0 -g + 5f = -13 g = 5f + 13 --- (2)

Substituting $g$ $g$ from (2) into (1):

2(5f + 13) + 3f = -13

13f + 26 = -13 f = -3

g = 5(-3) + 13 = -2

The circle equation is:

x² + y² - 4x - 6y = 0

Calculating radius squared $r²$ $r^2$:

r² = g² + f² - 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:

k² + (3k)² - 4k - 6(3k) = 0

10k² - 22k = 0 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 k = (11)/(5)

Now, calculate $10k + r²$ $10k + r^2$:

10k + r² = 10((11)/(5)) + 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.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 10 Mathematics: Coordinate Geometry

Q67 jee_main_2025_03_april_evening Ellipse

Let

C

be the circle of minimum area enclosing the ellipse

E: (x²)/(a²) + (y²)/(b²) = 1

with eccentricity

(1)/(2)

and foci

(pm 2, 0)

. Let

PQR

be a variable triangle, whose vertex

P

is on the circle

C

and the side

QR

of length

8

is parallel to the major axis of

E

and contains the point of intersection of

E

with the negative

y

-axis. Then the maximum area of the triangle

PQR

is:

A. $6(3 + \sqrt(2))$ $6(3 + \sqrt{2})$
B. $8(3 + \sqrt(2))$ $8(3 + \sqrt{2})$
C. $6(2 + \sqrt(3))$ $6(2 + \sqrt{3})$
D. $8(2 + \sqrt(3))$ $8(2 + \sqrt{3})$

Solution

Related Formula

For an ellipse $E$ $E$:

Foci: $(\pm ae, 0)$ $(\pm ae, 0)$
Eccentricity: $b² = a²(1 - e²)$ $b^2 = a^2(1 - e^2)$
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: $Area = (1)/(2) \cdot base \cdot height$

Core Logic

Let's first find coordinates $a$ $a$ and $b$ $b$:

$ae = 2$ $ae = 2$
$e = (1)/(2) a((1)/(2)) = 2 a = 4$
$b² = a²(1 - e²) = 16(1 - (1)/(4)) = 12 b = 2\sqrt(3)$

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:

x² + y² = 16 P = (4 θ, 4 θ)

The intersection of the ellipse with the negative $y$ $y$-axis is $(0, -b) = (0, -2\sqrt(3))$ $(0, -b) = (0, -2\sqrt{3})$.

Since side $QR$ $QR$ (length $= 8$ $= 8$) is parallel to the major axis ( $x$ $x$-axis) and contains $(0, -2\sqrt(3))$ $(0, -2\sqrt{3})$, the equation of the line containing $QR$ $QR$ is: $y = -2\sqrt(3)$ $y = -2\sqrt{3}$

Enclosing circle diagram for Q67 - JEE Main 2025 Evening Shift

Step 2: Maximizing Area of $Δ PQR$ $\Delta PQR$

The perpendicular height of vertex $P(4 θ, 4 θ)$ $P(4\cos\theta, 4\sin\theta)$ from the base line $y = -2\sqrt(3)$ $y = -2\sqrt{3}$ is:

H = 4 θ - (-2\sqrt(3)) = 4 θ + 2\sqrt(3)

To maximize the area, we maximize height $H$ $H$ by choosing $θ = 1$ $\sin\theta = 1$:

Hmax = 4 + 2\sqrt(3)

Maximum Area = (1)/(2) \cdot base QR \cdot Hmax

Maximum Area = (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 $Area = (1)/(2) \cdot base \cdot (a+b)$ .

Chapter Mix

Class 11 Mathematics: Conic Sections

Q68 jee_main_2025_03_april_evening Parabola

The shortest distance between the curves

y² = 8x

and

x² + y² + 12y + 35 = 0

is :

A. $2\sqrt(3) - 1$ $2\sqrt{3} - 1$
B. $\sqrt(2)$ $\sqrt{2}$
C. $3\sqrt(2) - 1$ $3\sqrt{2} - 1$
D. $2\sqrt(2) - 1$ $2\sqrt{2} - 1$

Solution

Related Formula

For a circle $x² + (y-k)² = R²$ $x^2 + (y-k)^2 = R^2$ and any smooth curve, the shortest distance lies along the normal to the curve passing through the center of the circle $C(h,k)$ $C(h,k)$:

Shortest Distance = 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:

x² + y² + 12y + 35 = 0 x² + (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 $y² = 8x$ $y^2 = 8x$, where $a = 2$ $a = 2$.

Normal equation of $y² = 4ax$ $y^2 = 4ax$ in slope form:

y = mx - 2am - am³

Substituting $a=2$ $a=2$:

y = mx - 4m - 2m³

Step 1: Find normal passing through circle center

Normal passes through $C(0, -6)$ $C(0, -6)$:

-6 = m(0) - 4m - 2m³

2m³ + 4m - 6 = 0 m³ + 2m - 3 = 0

By inspection, $m=1$ $m=1$ is a real solution:

(m-1)(m² + m + 3) = 0

Since $m² + m + 3 = 0$ $m^2 + m + 3 = 0$ 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 $P(am², -2am)$ $P(am^2, -2am)$ is:

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) = \sqrt(4 + 4) = 2\sqrt(2)

Shortest distance:

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.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q75 jee_main_2025_03_april_evening Hyperbola

If the equation of the hyperbola with foci

(4, 2)

and

(8, 2)

is

3x² - y² - α x + β y + γ = 0

, then

α + β + γ

is equal to

Numerical Answer. Answer: 141 to 141

Solution

Related Formula

For a horizontal hyperbola centered at $(h,k)$ $(h,k)$:

((x-h)²)/(a²) - ((y-k)²)/(b²) = 1

Foci: $(h \pm ae, k)$ $(h \pm ae, k)$
Eccentricity relation: $b² = a²(e² - 1) = a² e² - a²$ $b^2 = a^2(e^2 - 1) = a^2 e^2 - a^2$

Core Logic

Foci are $S₁ = (4,2)$ $S_1 = (4,2)$ and $S₂ = (8,2)$ $S_2 = (8,2)$.

Center $C(h,k)$ $C(h,k)$ is the midpoint:

h = (4 + 8)/(2) = 6, k = 2 C = (6, 2)

Distance between foci:

2ae = 8 - 4 = 4 ae = 2

Thus, $b² = 4 - a²$ $b^2 = 4 - a^2$.

Step 1: Expanding standard equation

The equation is:

((x-6)²)/(a²) - ((y-2)²)/(4-a²) = 1

(4-a²)(x-6)² - a²(y-2)² = a²(4-a²)

Comparing with $3x² - y² - α x + β y + γ = 0$ , the ratio of coefficients of $x²$ $x^2$ and $y²$ $y^2$ is $(3)/(-1) = -3$ $\frac{3}{-1} = -3$:

(4 - a²)/(-a²) = -3 4 - a² = 3a² 4a² = 4 a² = 1

Thus, $b² = 4 - 1 = 3$ $b^2 = 4 - 1 = 3$.

Step 2: Finding values of coefficients $α, β, γ$ $\alpha, \beta, \gamma$

Substituting $a² = 1$ $a^2 = 1$ back into standard form equation:

3(x-6)² - (y-2)² = 3

3(x² - 12x + 36) - (y² - 4y + 4) = 3

3x² - 36x + 108 - y² + 4y - 4 = 3

3x² - y² - 36x + 4y + 101 = 0

Comparing coefficients:

$α = 36$ $\alpha = 36$
$β = 4$ $\beta = 4$
$γ = 101$ $\gamma = 101$

α + β + γ = 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.

Chapter Mix

Class 11 Conic Sections

Q jee_main_2025_07_april_morning Parabola

Let

P

be the parabola, whose focus is

(-2, 1)

and directrix is

2x + y + 2 = 0

. Then the sum of the ordinates of the points on

P

, whose abscissa is

-2

, is

A. $(3)/(2)$ $\frac{3}{2}$
B. $(5)/(2)$ $\frac{5}{2}$
C. $(1)/(4)$ $\frac{1}{4}$
D. $(3)/(4)$ $\frac{3}{4}$

Solution

Related Formula

By the definition of a parabola, the distance from any point $(x, y)$ $(x, y)$ on the curve to the focus $(xf, yf)$ $(x_f, y_f)$ equals its perpendicular distance to the directrix line $Ax + By + C = 0$ $Ax + By + C = 0$:

(x - xf)² + (y - yf)² = ((Ax + By + C)²)/(A² + B²)

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 = ((2x + y + 2)²)/(2² + 1²)

5[(x + 2)² + (y - 1)²] = (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

. Substitute

x = -2

into the general equation:

5[(-2 + 2)² + (y - 1)²] = (2(-2) + y + 2)²

5[0 + (y - 1)²] = (-4 + y + 2)²

5(y - 1)² = (y - 2)²

5(y² - 2y + 1) = y² - 4y + 4

5y² - 10y + 5 = y² - 4y + 4

4y² - 6y + 1 = 0

Step 2: Find the Sum of Ordinates

The ordinates $y₁$ $y_1$ and $y₂$ $y_2$ are the roots of the quadratic equation $4y² - 6y + 1 = 0$ $4y^2 - 6y + 1 = 0$. The sum of the ordinates is:

y₁ + y₂ = -(-6)/(4) = (6)/(4) = (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.

Chapter Mix

Class 11 Mathematics: Conic Sections

A line passing through the point P(√(5), √(5))$P(\sqrt{5}, \sqrt{5})$ intersects the ellipse (x²)/(36) + (y²)/(25) = 1$\frac{x^2}{36} + \frac{y^2}{25} = 1$ at A$A$ and B$B$ [cite: 567] such that (PA) · (PB)$(PA) \cdot (PB)$ is maximum. Then 5(PA² + PB²)$5(PA^{2} + PB^{2})$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Maximization Condition

Step 2: Distance Value Summation

Pattern Recognition

Evaluation Rubric / Model Answer

Chapter Mix

More Conic Sections Previous-Year Questions — Page 8

Solution

Related Formula

Core Logic

Step 1: Finding g$g$, f$f$ and r²$r^2$

Step 2: Solving for k$k$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Setting Circle and Triangle geometry

Step 2: Maximizing Area of Δ PQR$\Delta PQR$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Find normal passing through circle center

Step 2: Point calculation and shortest distance

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Expanding standard equation

Step 2: Finding values of coefficients α, β, γ$\alpha, \beta, \gamma$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute the Given Abscissa

Step 2: Find the Sum of Ordinates

Pattern Recognition

Chapter Mix

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Step 1: Finding $g$ $g$, $f$ $f$ and $r²$ $r^2$

Step 2: Solving for $k$ $k$

Step 2: Maximizing Area of $Δ PQR$ $\Delta PQR$

Step 2: Finding values of coefficients $α, β, γ$ $\alpha, \beta, \gamma$