Let ABCD be a trapezium whose vertices lie on the parabola y2 = 4x. Let the sides AD and BC of the trapezium be…

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Q64 jee_main_2025_04_april_evening Ellipse

Let for two distinct values of

p

the lines

y = x + p

touch the ellipse

E: (x²)/(4²) + (y²)/(3²) = 1

at the points A and B. Let the line

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 $(x²)/(a²) + (y²)/(b²) = 1$ $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is:

p² = a²m² + b²

The coordinate of the point of contact is given by $(-(a²m)/(p), (b²)/(p))$ .

Core Logic

Given the ellipse parameter values $a² = 16$ $a^2 = 16$ and $b² = 9$ $b^2 = 9$, and tangent line slope $m = 1$ $m = 1$:

p² = 16(1)² + 9 = 25 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 $p = 5$ $p = 5$: $A = (-(16(1))/(5), (9)/(5)) = (-(16)/(5), (9)/(5))$
For $p = -5$ $p = -5$: $B = (-(16(1))/(-5), (9)/(-5)) = ((16)/(5), -(9)/(5))$

Step 1: Intersecting line with Ellipse

The line $y = x$ $y = x$ intersects the ellipse $(x²)/(16) + (y²)/(9) = 1$ $\frac{x^2}{16} + \frac{y^2}{9} = 1$:

(x²)/(16) + (x²)/(9) = 1 (25x²)/(144) = 1 x² = (144)/(25) x = \pm (12)/(5)

Since $y = x$ $y = x$, the intersection points $C$ $C$ and $D$ $D$ are:

C = (-(12)/(5), -(12)/(5)) and D = ((12)/(5), (12)/(5))

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:

Area = (1)/(2) vmatrix xA & yA & 1 xB & yB & 1 xC & yC & 1 vmatrix + = 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 $ABD$ $ABD$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q67 jee_main_2025_04_april_morning Ellipse - Foci and Latus Rectum

The length of the latus-rectum of the ellipse, whose foci are

(2, 5)

and

(2, -3)

and eccentricity is

(4)/(5)

, is

A. $(6)/(5)$ $\frac{6}{5}$
B. $(50)/(3)$ $\frac{50}{3}$
C. $(10)/(3)$ $\frac{10}{3}$
D. $(18)/(5)$ $\frac{18}{5}$

Solution

Related Formula

Distance between foci $= 2be$ $= 2be$ (for vertical major axis). Length of Latus Rectum $(L.R.) = (2a²)/(b)$ $(\text{L.R.}) = \frac{2a^2}{b}$.

Core Logic

Foci are $F₁(2,5)$ $F_1(2,5)$ and $F₂(2,-3)$ $F_2(2,-3)$. The $x$ $x$-coordinates are identical, indicating a vertical ellipse. Distance between foci:

2be = 5 - (-3) = 8 be = 4

Given eccentricity $e = (4)/(5)$ $e = \frac{4}{5}$:

b((4)/(5)) = 4 b = 5

Ellipse - Foci and Latus Rectum diagram for Q67 - JEE Main 2025 Morning

Step 1: Calculate Minor Axis length

Use eccentricity relation:

a² = b²(1 - e²) a² = 25(1 - (16)/(25)) = 25 \times (9)/(25) = 9 a = 3

Step 2: Evaluate Latus Rectum

L.R. = (2a²)/(b) = (2 \times 9)/(5) = (18)/(5)

Pattern Recognition

Always verify axis orientation (horizontal vs vertical) from coordinates before blindly substituting into standard formulas. Symmetrical components match axis lengths parameters directly.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q74 jee_main_2025_04_april_morning Tangents to Conics

Let C be the circle

x² + (y - 1)² = 2

,

E₁

and

E₂

be two ellipses whose centres lie at the origin and major axes lie on the x-axis and y-axis respectively. Let the straight line

x + y = 3

touch the curves C,

E₁

and

E₂

at

P(x₁, y₁)

,

Q(x₂, y₂)

and

R(x₃, y₃)

respectively. Given that P is the mid-point of the line segment QR and

PQ = 2\sqrt(2)3

, the value of

9(x₁y₁ + x₂y₂ + x₃y₃)

is equal to

Numerical Answer. Answer: 46 to 46

Solution

Related Formula

Parametric equation of a straight line:

x = x₁ + r θ, y = y₁ + r θ

Core Logic

Step 1: Find point $P(x₁,y₁)$ $P(x_1,y_1)$ on circle $C$ $C$. Equation of tangent at $P$ $P$ on $x² + y² - 2y - 1 = 0$ $x^2 + y^2 - 2y - 1 = 0$ is $xx₁ + y(y₁ - 1) - (y₁ + 1) = 0$ $xx_1 + y(y_1 - 1) - (y_1 + 1) = 0$. Comparing with line $x + y = 3 (x₁)/(1) = (y₁ - 1)/(1) = (y₁ + 1)/(3)$ . Solving gives $x₁ = 1, y₁ = 2$ $x_1 = 1, y_1 = 2$. Thus, $P = (1, 2)$ $P = (1, 2)$.

Step 1: Use Line Parametrics for Q and R

Line $x + y = 3$ $x + y = 3$ makes an angle $θ = 135°$ $\theta = 135^{\circ}$ with the positive x-axis. Using parametric distances from $P(1,2)$ $P(1,2)$ with $r = PQ = 2\sqrt(2)3$ $r = PQ = \frac{2\sqrt{2}}{3}$:

x = 1 \pm r (135°) = 1 \mp r\sqrt(2)

y = 2 \pm r (135°) = 2 \pm r\sqrt(2)

Substitute $r = 2\sqrt(2)3$ $r = \frac{2\sqrt{2}}{3}$: For $Q$ $Q$: $x₂ = 1 + (2)/(3) = (5)/(3), y₂ = 2 - (2)/(3) = (4)/(3)$ . For $R$ $R$: $x₃ = 1 - (2)/(3) = (1)/(3), y₃ = 2 + (2)/(3) = (8)/(3)$ .

Step 2: Evaluate Final Expression

Calculate the products: $x₁y₁ = 1 \times 2 = 2$ $x_1y_1 = 1 \times 2 = 2$ $x₂y₂ = (5)/(3) \times (4)/(3) = (20)/(9)$ $x₃y₃ = (1)/(3) \times (8)/(3) = (8)/(9)$

9(x₁y₁ + x₂y₂ + x₃y₃) = 9(2 + (20)/(9) + (8)/(9)) = 18 + 20 + 8 = 46

Pattern Recognition

Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations $a², b²$ $a^2, b^2$ completely.

Chapter Mix

Class 11 Mathematics: Circles Class 11 Mathematics: Conic Sections

Q62 jee_main_2025_07_april_evening Properties of Ellipse

Let the length of a latus rectum of an ellipse

(x²)/(a²) + (y²)/(b²) = 1

be 10. If its eccentricity is the minimum value of the function

f(t) = t² + t + (11)/(12), t in R

, then

a² + b²

is equal to:

A. $125$ $125$
B. $126$ $126$
C. $120$ $120$
D. $115$ $115$

Solution

Related Formula

Length of latus rectum of an ellipse and its eccentricity relation are:

LR = (2b²)/(a)

e² = 1 - (b²)/(a²)

Core Logic

Given length of $LR = 10 (2b²)/(a) = 10 b² = 5a (i)$

Now, let's find the minimum value of $f(t) = t² + t + (11)/(12)$ $f(t) = t^2 + t + \frac{11}{12}$. Differentiating: $f'(t) = 2t + 1 = 0 t = -(1)/(2)$ .

Minimum value e = f(-(1)/(2)) = (-(1)/(2))² + (-(1)/(2)) + (11)/(12) = (1)/(4) - (1)/(2) + (11)/(12) = (3 - 6 + 11)/(12) = (8)/(12) = (2)/(3)

Step 1: Solve for a and b

Using eccentricity formula:

e² = (4)/(9) = 1 - (b²)/(a²) (b²)/(a²) = (5)/(9) b² = (5a²)/(9) (ii)

Equating (i) and (ii):

5a = (5a²)/(9) a = 9

Then from (i):

b² = 5(9) = 45 b = 3\sqrt(5)

Hence, $a² = 81$ $a^2 = 81$.

Step 2: Calculate a^2 + b^2

a² + b² = 81 + 45 = 126

Pattern Recognition

A quadratic function $at²+bt+c$ $at^2+bt+c$ reaches its extreme value at $t = -(b)/(2a)$ $t = -\frac{b}{2a}$. Using this layout avoids full calculus derivation steps.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

Q68 jee_main_2025_07_april_evening Eccentricity and Foci

Let

e₁

and

e₂

be the eccentricities of the ellipse

x²b² + y²25 = 1

and the hyperbola

x²16 - y²b² = 1

, respectively. If

b < 5

and

e₁e₂ = 1

, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is:

A. $(4)/(5)$ $\frac{4}{5}$
B. $(3)/(5)$ $\frac{3}{5}$
C. $\sqrt(7)4$ $\frac{\sqrt{7}}{4}$
D. $\sqrt(3)2$ $\frac{\sqrt{3}}{2}$

Solution

Related Formula

Eccentricity for ellipse ( $a $a) and hyperbola are calculated as:$

e₁² = 1 - (a²)/(b²)

e₂² = 1 + (b²)/(a²)

Core Logic

Since $b < 5$ $b < 5$, for the ellipse $(x²)/(b²) + (y²)/(25) = 1$ $\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the major axis is along the $y$ $y$-axis:

e₁² = 1 - (b²)/(25)

For the hyperbola $(x²)/(16) - (y²)/(b²) = 1$ $\frac{x^2}{16} - \frac{y^2}{b^2} = 1$:

e₂² = 1 + (b²)/(16)

Given $e₁² e₂² = 1$ $e_1^2 e_2^2 = 1$:

(1 - (b²)/(25))(1 + (b²)/(16)) = 1

1 + (b²)/(16) - (b²)/(25) - (b⁴)/(400) = 1 (9b²)/(400) = (b⁴)/(400) b² = 9

Step 1: Calculate Foci Locations

Substituting $b² = 9$ $b^2 = 9$: Ellipse foci: $ae₁ = 5 \cdot \sqrt(1 - (9)/(25)) = 5 \cdot (4)/(5) = 4$ . Foci lie along $y$ $y$-axis: $(0, \pm 4)$ $(0, \pm 4)$. Hyperbola foci: $ae₂ = 4 \cdot \sqrt(1 + (9)/(16)) = 4 \cdot (5)/(4) = 5$ . Foci lie along $x$ $x$-axis: $(\pm 5, 0)$ $(\pm 5, 0)$.

Step 2: Construct the New Ellipse

The new ellipse passes through $(0, \pm 4)$ $(0, \pm 4)$ and $(\pm 5, 0)$ $(\pm 5, 0)$. Thus, its semi-major axis is $A = 5$ $A = 5$ along the $x$ $x$-axis and semi-minor axis is $B = 4$ $B = 4$ along the $y$ $y$-axis:

E = \sqrt(1 - (B²)/(A²)) = \sqrt(1 - (16)/(25)) = (3)/(5)

Pattern Recognition

When a conic passes through points directly on the axes like $(\pm alpha, 0)$ $(\pm alpha, 0)$ and $(0, \pm β)$ $(0, \pm \beta)$, those points immediately represent the semi-axes values $A$ $A$ and $B$ $B$.

Chapter Mix

Class 11 Mathematics: Conic Sections

Let ABCD be a trapezium whose vertices lie on the parabola y² = 4x$y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length (25)/(4)$\frac{25}{4}$ and it passes through the point (1,0)$(1,0)$, then the area of ABCD is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Using Diagonal Properties

Step 2: Finding Coordinates and Area

Pattern Recognition

Chapter Mix

More Conic Sections Previous-Year Questions — Page 10

Solution

Related Formula

Core Logic

Step 1: Intersecting line with Ellipse

Step 2: Calculating Quadrilateral Area

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Minor Axis length

Step 2: Evaluate Latus Rectum

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Use Line Parametrics for Q and R

Step 2: Evaluate Final Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Solve for a and b

Step 2: Calculate a^2 + b^2

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Foci Locations

Step 2: Construct the New Ellipse

Pattern Recognition

Chapter Mix

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Let ABCD be a trapezium whose vertices lie on the parabola $y² = 4x$ $y^2 = 4x$. Let the sides AD and BC of the trapezium be parallel to y-axis. If the diagonal AC is of length $(25)/(4)$ $\frac{25}{4}$ and it passes through the point $(1,0)$ $(1,0)$, then the area of ABCD is: