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

Q14 jee_main_2024_30_jan_morning Intersection of Two Circles

If the circles

(x + 1)^2 + (y + 2)^2 = r^2

and

x^2 + y^2 - 4x - 4y + 4 = 0

intersect at exactly two distinct points, then

A. $5 < r < 9$ $5 < r < 9$
B. $0 < r < 7$ $0 < r < 7$
C. $3 < r < 7$ $3 < r < 7$
D. $frac12 < r < 7$ $\frac{1}{2} < r < 7$

Solution

### Related Formula

|r_1 - r_2| < C_1 C_2 < r_1 + r_2

where

C_1 C_2

is the distance between the centers and

r_1, r_2

are the radii. ### Core Logic Circle 1:

(x + 1)^2 + (y + 2)^2 = r^2

Center

C_1(-1, -2)

, Radius

r_1 = r

Circle 2:

x^2 + y^2 - 4x - 4y + 4 = 0

Convert to standard form:

(x - 2)^2 - 4 + (y - 2)^2 - 4 + 4 = 0 Rightarrow (x - 2)^2 + (y - 2)^2 = 4

Center

C_2(2, 2)

, Radius

r_2 = 2

### Step 1: Distance between centers Distance

C_1C_2 = sqrt(2 - (-1))^2 + (2 - (-2))^2 = sqrt3^2 + 4^2 = sqrt9 + 16 = 5

. ### Step 2: Applying inequality condition For exactly two intersection points:

|r_1 - r_2| < C_1 C_2 < r_1 + r_2

|r - 2| < 5 < r + 2

Part A:

r + 2 > 5 Rightarrow r > 3 quad dots (1)

Part B:

|r - 2| < 5

-5 < r - 2 < 5

-3 < r < 7 quad dots (2)

### Step 3: Intersection of inequalities Taking the intersection of (1) and (2):

r in (3, infty) cap (-3, 7) = (3, 7)

3 < r < 7

### Pattern Recognition Intersecting circles always strictly obey the triangle inequality formed by their radii and the distance between their centers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

Q15 jee_main_2024_30_jan_morning Ellipse Eccentricity

If the length of the minor axis of ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is:

A. $fracsqrt53$ $\frac{\sqrt{5}}{3}$
B. $fracsqrt32$ $\frac{\sqrt{3}}{2}$
C. $frac1sqrt3$ $\frac{1}{\sqrt{3}}$
D. $frac2sqrt5$ $\frac{2}{\sqrt{5}}$

Solution

### Related Formula

b^2 = a^2(1 - e^2)

Length of minor axis

= 2b

Distance between foci

= 2ae

### Core Logic Given condition: Length of minor axis = Half of the distance between foci.

2b = frac12(2ae)

2b = ae

fracba = frace2

### Step 1: Squaring and finding eccentricity Squaring both sides:

fracb^2a^2 = frace^24

We know

fracb^2a^2 = 1 - e^2

.

1 - e^2 = frace^24

1 = e^2 + frace^24 = frac5e^24

e^2 = frac45

e = frac2sqrt5

### Pattern Recognition Proportional relationships between

a, b,

and

ae

in an ellipse can be immediately plugged into the fundamental identity

b^2 = a^2(1-e^2)

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

Q23 jee_main_2024_30_jan_morning Hyperbola

Let the latus rectum of the hyperbola

fracx^29 - fracy^2b^2 = 1

subtend an angle of

fracpi3

at the centre of the hyperbola. If

b^2

is equal to

fraclm(1 + sqrtn)

, where

l

and

m

are co-prime numbers, then

l^2 + m^2 + n^2

is equal to

Numerical Answer. Answer: 182 to 182

Solution

### Related Formula

Latus Rectum = frac2b^2a

e^2 = 1 + fracb^2a^2

### Core Logic

Hyperbola diagram for Q23 - JEE Main 2024 Morning

The endpoints of the latus rectum are

(ae, fracb^2a)

and

(ae, -fracb^2a)

. It subtends

60^circ

at the center

(0,0)

. By symmetry, the line connecting the center to

(ae, fracb^2a)

makes an angle of

30^circ

with the x-axis.

tan 30^circ = fracfracb^2aae = fracb^2a^2e = frac1sqrt3

### Step 1: Expressing e in terms of b Given

a^2 = 9

, so

a = 3

.

fracb^29e = frac1sqrt3 Rightarrow e = fracsqrt3b^29

### Step 2: Applying eccentricity relation We know

e^2 = 1 + fracb^29

. Substitute

e

:

left(fracsqrt3b^29right)^2 = 1 + fracb^29

frac3b^481 = 1 + fracb^29

fracb^427 = frac9 + b^29

b^4 = 3(9 + b^2) = 27 + 3b^2

b^4 - 3b^2 - 27 = 0

### Step 3: Solving for b^2 Using the quadratic formula for

b^2

:

b^2 = frac3 pm sqrt9 - 4(1)(-27)2 = frac3 pm sqrt9 + 1082 = frac3 pm sqrt1172

Since

b^2 > 0

:

b^2 = frac3 + 3sqrt132 = frac32(1 + sqrt13)

Comparing this to

fraclm(1 + sqrtn)

:

l = 3

,

m = 2

,

n = 13

. ### Step 4: Final calculation Target:

l^2 + m^2 + n^2

= 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = 182

### Pattern Recognition Whenever an angle is subtended at the origin symmetrically, use half the angle mapping to the tangent of the coordinate ratio

(y/x)

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

Q7 jee_main_2024_31_jan_evening Parabola and Ellipse intersection

Let

P

be a parabola with vertex

(2, 3)

and directrix

2x + y = 6

. Let an ellipse

E:fracx^2a^2 +fracy^2b^2 = 1,a > b

of eccentricity

frac1sqrt2

pass through the focus of the parabola

P

. Then the square of the length of the latus rectum of

E

, is

A. $frac3858$ $\frac{385}{8}$
B. $frac3478$ $\frac{347}{8}$
C. $frac51225$ $\frac{512}{25}$
D. $frac65625$ $\frac{656}{25}$

Solution

### Related Formula

textLength of Latus Rectum of Ellipse = frac2b^2a

e = sqrt1 - fracb^2a^2

### Core Logic

Parabola and Ellipse intersection diagram for Q7 - JEE Main 2024 Evening

Directrix of parabola is

2x + y - 6 = 0

. Axis is perpendicular to directrix and passes through vertex

V(2,3)

. Slope of axis

= 1/2

. Equation of axis:

y - 3 = frac12(x - 2) implies x - 2y + 4 = 0

. Intersection of axis and directrix is

K

:

2x + y = 6

x - 2y = -4

Solving,

K = (8/5, 14/5) = (1.6, 2.8)

. Vertex

V(2,3)

is the midpoint of focus

S(alpha, beta)

and

K(1.6, 2.8)

:

fracalpha + 1.62 = 2 implies alpha = 2.4

fracbeta + 2.82 = 3 implies beta = 3.2

So, focus

S = (2.4, 3.2)

. Ellipse passes through

S(2.4, 3.2)

:

frac(2.4)^2a^2 + frac(3.2)^2b^2 = 1

Eccentricity

e = frac1sqrt2 implies 1 - fracb^2a^2 = frac12 implies a^2 = 2b^2

. Substitute

a^2 = 2b^2

:

frac5.762b^2 + frac10.24b^2 = 1 implies frac2.88 + 10.24b^2 = 1 implies b^2 = 13.12 = frac32825

Square of Latus Rectum:

L^2 = left(frac2b^2aright)^2 = frac4b^4a^2 = frac4b^42b^2 = 2b^2 = 2 times frac32825 = frac65625

### Pattern Recognition Leverage geometry of parabola (Vertex is exactly midway between focus and directrix intersection on the axis) to find coordinate points before substituting into ellipse equation. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

Q3 jee_main_2024_31_jan_morning Hyperbola and Ellipse Properties

If the foci of a hyperbola are same as that of the ellipse

fracx^29 + fracy^225 = 1

and the eccentricity of the hyperbola is

frac158

times the eccentricity of the ellipse, then the smaller focal distance of the point

left(sqrt2, frac143sqrtfrac25right)

on the hyperbola, is equal to

A. $7sqrtfrac25 - frac83$ $$7\sqrt{\frac{2}{5}} - \frac{8}{3}$$
B. $14sqrtfrac25 - frac43$ $$14\sqrt{\frac{2}{5}} - \frac{4}{3}$$
C. $14sqrtfrac25 - frac163$ $$14\sqrt{\frac{2}{5}} - \frac{16}{3}$$
D. $7sqrtfrac25 + frac83$ $$7\sqrt{\frac{2}{5}} + \frac{8}{3}$$

Solution

### Related Formula

e = sqrt1 - fraca^2b^2 text (for vertical ellipse)

textFocal distance of P(x_1, y_1) text on hyperbola = e_H |y_1 pm fracBe_H|

### Core Logic For the ellipse

fracx^29 + fracy^225 = 1

:

a = 3, b = 5

. Since

b > a

, the major axis is along the y-axis.

e = sqrt1 - frac925 = frac45

Foci

= (0, pm be) = (0, pm 4)

. ### Step 1: Hyperbola Properties Eccentricity of hyperbola

e_H = frac45 times frac158 = frac32

. Let the hyperbola be

fracx^2A^2 - fracy^2B^2 = -1

since its foci

(0, pm 4)

are on the y-axis. Foci

B e_H = 4 implies B = frac83

.

A^2 = B^2(e_H^2 - 1) = frac649left(frac94 - 1right) = frac809

. Equation:

fracx^280/9 - fracy^264/9 = -1

. ### Step 2: Focal Distance Calculation Directrix of hyperbola:

y = pm fracBe_H = pm frac169

. Smaller focal distance

PS = e_H cdot PM

, where

PM

is the distance to the nearer directrix.

PS = frac32 left| frac143sqrtfrac25 - frac169 right| = 7sqrtfrac25 - frac83

### Pattern Recognition Smaller focal distance of a point

(x_1, y_1)

on a vertical hyperbola is given by

e|y_1 - fracbe|

where

y_1 > 0

and we use the positive directrix. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Maths: Conic Sections

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 $y^2 = 9x$ $y^2 = 9x$ at the point $(4, 6)$ $(4, 6)$ . Then $r$ $r$ is equal to

Solution & Explanation

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 11

Solution

Solution

Solution

Solution

Solution

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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 y^2 = 9x$y^2 = 9x$ at the point (4, 6)$(4, 6)$ . Then r$r$ is equal to

Solution & Explanation

Reference Study Guides

More Conic Sections Previous-Year Questions — Page 11

Solution

Solution

Solution

Solution

Solution

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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 $y^2 = 9x$ $y^2 = 9x$ at the point $(4, 6)$ $(4, 6)$ . Then $r$ $r$ is equal to