Let r be the radius of the circle, which touches x -axis at point (a, 0) , a

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More Conic Sections Previous-Year Questions — Page 7

Q75 jee_main_2025_03_april_evening Hyperbola

If the equation of the hyperbola with foci

(4, 2)

and

(8, 2)

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

Q73 jee_main_2025_07_april_morning Hyperbola

Consider the hyperbola

(x²)/(a²) -(y²)/(b²) = 1

having one of its focus at

P(-3,0)

. If the latus rectum through its other focus subtends a right angle at

P

and

a² b² = α \sqrt(2) -β ,α ,β in N

, calculate

α + β

Numerical Answer. Answer: 1944 to 1944

Solution

Related Formula

For a standard hyperbola:

Focus positions are $(\pm ae, 0)$ $(\pm ae, 0)$.
Length of semi-latus rectum is $(b²)/(a)$ $\frac{b^2}{a}$.
Eccentricity identity linkage: $b² = a²(e² - 1) a²e² = a² + b²$ .

Core Logic

Given focus $F₁ \equiv (-ae, 0) \equiv P(-3, 0)$ $F_1 \equiv (-ae, 0) \equiv P(-3, 0)$, so $ae = 3$ $ae = 3$. The other focus is $F₂ \equiv (ae, 0) \equiv (3, 0)$ $F_2 \equiv (ae, 0) \equiv (3, 0)$.

The latus rectum passes vertically through $F₂$ $F_2$, with endpoints $L₁(ae, (b²)/(a))$ $L_1\left(ae, \frac{b^2}{a}\right)$ and $L₂(ae, -(b²)/(a))$ $L_2\left(ae, -\frac{b^2}{a}\right)$. This segment subtends a right angle at $P(-ae, 0)$ $P(-ae, 0)$. By symmetry, the top half angle at $P$ $P$ must be exactly $45^°$ $45^\circ$.

Step 1: Set Up Slope Relationship

Hyperbola diagram for Q73 - JEE Main 2025 Morning

Using the geometric slope relationship:

45^° = heightbase = (b²/a)/(2ae)

1 = (b²)/(2a²e) 2a²e = b² b² = 6a (since ae = 3)

Step 2: Solve the Quadratic Excentricity Equation

Substitute $ae = 3$ $ae = 3$ and $b² = 6a$ $b^2 = 6a$ into the eccentricity identity $a²e² = a² + b²$ $a^2e^2 = a^2 + b^2$:

9 = a² + 6a a² + 6a - 9 = 0

Solving for $a$ $a$ using the quadratic formula (taking the positive root since $a > 0$ $a > 0$):

a = -6 \pm \sqrt(36 - 4(1)(-9))2 = -6 + \sqrt(72)2 = -3 + 3\sqrt(2) = 3(\sqrt(2) - 1)

Step 3: Evaluate product and sum coefficients

Now compute $a²b²$ $a^2b^2$:

a²b² = a²(6a) = 6a³

6a³ = 6[3(\sqrt(2) - 1)]³ = 6 \times 27 \times (\sqrt(2) - 1)³

6a³ = 162 \times (2\sqrt(2) - 6 + 3\sqrt(2) - 1) = 162 \times (5\sqrt(2) - 7)

6a³ = 810\sqrt(2) - 1134

Matching with $α\sqrt(2) - β$ $\alpha\sqrt{2} - \beta$ gives:

α = 810 and β = 1134

Calculate the final required sum:

α + β = 810 + 1134 = 1944

Pattern Recognition

Recognizing that the right angle subtended at the opposite focus implies a perfect $(45^°)$ $\tan(45^\circ)$ right triangle instantly yields the key linear constraint $b² = 2a(ae)$ $b^2 = 2a(ae)$, avoiding the need for lengthy distance-formula tracking.

Chapter Mix

Class 11 Mathematics: Conic Sections

Q60 jee_main_2025_08_april_evening Ellipse and Focal Distances

Let the ellipse

3x² + py² = 4

pass through the centre C of the circle

x² + y² - 2x - 4y - 11 = 0

of radius r. Let

f₁, f₂

be the focal distances of the point C on the ellipse. Then

6f₁f₂ - r

is equal to

A. $74$ $74$
B. $68$ $68$
C. $70$ $70$
D. $78$ $78$

Solution

Related Formula

Focal Distance Product on Vertical Ellipse = b² - e² k²

Core Logic

Extract the coordinate center of the target circle, substitute it directly to locate the missing parameter $p$ $p$, and resolve eccentricity metrics.

Step 1: Extract Circle Metric Values

For circle $x² + y² - 2x - 4y - 11 = 0$ $x^{2} + y^{2} - 2x - 4y - 11 = 0$:

Centre C(1, 2), Radius r = \sqrt(1 + 4 + 11) = 4

Step 2: Standardize Ellipse Formulation

Ellipse passes through point $C(1,2)$ $C(1,2)$:

3(1)² + p(2)² = 4 3 + 4p = 4 p = (1)/(4)

Standard model form: $(x²)/(4/3) + (y²)/(16) = 1$ $\frac{x^2}{4/3} + \frac{y^2}{16} = 1$ ( $b > a$ $b > a$, vertical configuration axis).

e = \sqrt(1 - (4/3)/(16)) = \sqrt(1 - (1)/(12)) = \sqrt((11)/(12))

Step 3: Evaluate Product Chain

Focal distance elements at ordinate coordinate height $k=2$ $k=2$ are bounded by $b \pm ek$ $b \pm ek$:

f₁ f₂ = b² - e² k² = 16 - ((11)/(12)) \times 4 = 16 - (11)/(3) = (37)/(3)

Target evaluation expression response string:

6f₁ f₂ - r = 6 ((37)/(3)) - 4 = 74 - 4 = 70

Pattern Recognition

Pay attention to whether $b > a$ $b > a$ or $a > b$ $a > b$ when analyzing ellipse forms. Focal distance definitions swap directions immediately across major horizontal/vertical configurations.

Chapter Mix

Class 11 Mathematics: Conic Sections Class 11 Mathematics: Circles

Let r$r$ be the radius of the circle, which touches x$x$ -axis at point (a, 0)$(a, 0)$ , a < 0$a < 0$ and the parabola y² = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Derive Shared Parabola Tangent Line

Step 2: Build Geometric Metric Connections

Step 3: Solve for Radius Matrix Bounds

Pattern Recognition

Chapter Mix

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 7

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

Solution

Related Formula

Core Logic

Step 1: Set Up Slope Relationship

Step 2: Solve the Quadratic Excentricity Equation

Step 3: Evaluate product and sum coefficients

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Extract Circle Metric Values

Step 2: Standardize Ellipse Formulation

Step 3: Evaluate Product Chain

Pattern Recognition

Chapter Mix

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Let $r$ $r$ be the radius of the circle, which touches $x$ $x$ -axis at point $(a, 0)$ $(a, 0)$ , $a < 0$ $a < 0$ and the parabola $y² = 9x$ $y^2 = 9x$ at the point $(4, 6)$ $(4, 6)$ . Then $r$ $r$ is equal to

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