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

Q55 jee_main_2025_28_jan_evening Chord of Ellipse

If the midpoint of a chord of the ellipse

fracx^29+fracy^24=1

is

(sqrt2,4/3)

, and the length of the chord is

frac2sqrtalpha3

, then

alpha

is:

A. $18$ $18$
B. $22$ $22$
C. $26$ $26$
D. $20$ $20$

Solution

### Related Formula Equation of a chord of an ellipse with given midpoint

(x_1, y_1)

is given by

T = S_1

:

fracxx_1a^2 + fracyy_1b^2 = fracx_1^2a^2 + fracy_1^2b^2

### Core Logic Given ellipse:

fracx^29 + fracy^24 = 1

and midpoint

(x_1, y_1) = (sqrt2, frac43)

. Applying

T = S_1

:

fracxsqrt29 + fracy(4/3)4 = frac(sqrt2)^29 + frac(4/3)^24

fracsqrt2x9 + fracy3 = frac29 + frac1636

fracsqrt2x9 + fracy3 = frac29 + frac49 = frac69

Multiplying through by 9:

sqrt2x + 3y = 6 implies 3y = 6 - sqrt2x

### Step 1: Find Intersection Points with Ellipse Substitute

3y = 6 - sqrt2x

into the multiplied form of ellipse

4x^2 + 9y^2 = 36

:

4x^2 + (3y)^2 = 36

4x^2 + (6 - sqrt2x)^2 = 36

4x^2 + 36 + 2x^2 - 12sqrt2x = 36

6x^2 - 12sqrt2x = 0

6x(x - 2sqrt2) = 0

Thus,

x = 0

or

x = 2sqrt2

. ### Step 2: Find y-coordinates and Chord Length If

x_1 = 0 implies 3y_1 = 6 implies y_1 = 2

If

x_2 = 2sqrt2 implies 3y_2 = 6 - sqrt2(2sqrt2) = 6 - 4 = 2 implies y_2 = frac23

The end points of the chord are

A(0, 2)

and

B(2sqrt2, frac23)

.

textLength of chord AB = sqrt(2sqrt2 - 0)^2 + left(frac23 - 2right)^2

AB = sqrt8 + left(-frac43right)^2 = sqrt8 + frac169 = sqrtfrac889 = fracsqrt4 times 223 = frac2sqrt223

Comparing with

frac2sqrtalpha3

, we get

alpha = 22

. ### Pattern Recognition When the intersection equation results in a simple factoring like

6x^2 - 12sqrt2x = 0

, calculating the explicit coordinates is incredibly fast compared to using general formula roots equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Ellipse)

Q69 jee_main_2025_28_jan_evening Intersection of Curves and Locus of Centroid

If A and B are the points of intersection of the circle

x^2+y^2-8x=0

and the hyperbola

fracx^29-fracy^24=1

and a point P moves on the line

2x-3y+4=0

, then the centroid of

Delta PAB

lies on the line:

A. $4x-9y=12$ $4x-9y=12$
B. $x+9y=36$ $x+9y=36$
C. $9x-9y=32$ $9x-9y=32$
D. $6x-9y=20$ $6x-9y=20$

Solution

### Related Formula Centroid

(h, k)

of a triangle with vertices

(x_1,y_1), (x_2,y_2), (x_3,y_3)

:

h = fracx_1 + x_2 + x_33, quad k = fracy_1 + y_2 + y_33

### Core Logic Given equations: 1) Circle:

y^2 = 8x - x^2

2) Hyperbola:

4x^2 - 9y^2 = 36

Substitute circle's

y^2

into hyperbola equation:

4x^2 - 9(8x - x^2) = 36 implies 4x^2 - 72x + 9x^2 = 36

13x^2 - 72x - 36 = 0 implies (13x + 6)(x - 6) = 0

If

x = -6/13

,

y^2 < 0

(rejected). Thus,

x = 6

. Substituting

x = 6

into circle:

y^2 = 8(6) - 6^2 = 48 - 36 = 12 implies y = pm sqrt12

. The intersection points are

A(6, sqrt12)

and

B(6, -sqrt12)

. ### Step 1: Relate Centroid coordinates to P Let point

P

have coordinates

(alpha, beta)

. Since

P

lies on

2x - 3y + 4 = 0

:

2alpha - 3beta + 4 = 0 implies beta = frac2alpha + 43

Let the centroid be

(h, k)

:

h = frac6 + 6 + alpha3 = frac12 + alpha3 implies alpha = 3h - 12

k = fracsqrt12 - sqrt12 + beta3 = fracbeta3 implies beta = 3k

### Step 2: Form the Locus Equation Substitute

alpha

and \beta into the line equation of

P

:

2(3h - 12) - 3(3k) + 4 = 0

6h - 24 - 9k + 4 = 0

6h - 9k = 20

Replacing

(h, k)

with general coordinates

(x, y)

gives the locus:

6x - 9y = 20

### Pattern Recognition Notice how the

y

-coordinates of intersection points

A

and

B

are symmetric (

\,pmsqrt12\,

), meaning their sum is zero. This simplifies the expression for

k

instantly to just

beta/3

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Coordinate Geometry Class 11 Mathematics: Conic Sections

Q74 jee_main_2025_28_jan_evening Parabola Transformation and Geometric Properties

Let A and B be the two points of intersection of the line

y+5=0

and the mirror image of the parabola

y^2=4x

with respect to the line

x+y+4=0

. If d denotes the distance between A and B, and a denotes the area of

Delta SAB

, where S is the focus of the parabola

y^2=4x,

then the value of

(a+d)

is

Numerical Answer. Answer: 14 to 14

Solution

### Related Formula Mirror image mapping of point

(x, y)

about line

x+y+c = 0

is:

x' = -y - c, quad y' = -x - c

### Core Logic Instead of reflecting the entire curve, we can reflect the line

y + 5 = 0

across the line

x + y + 4 = 0

to find where it intersects the original parabola

y^2 = 4x

. Reflection of line

y = -5

across

x + y + 4 = 0

: Using transformation

y' = -x - 4 implies -5 = -x - 4 implies x = 1

. So the reflected line is

x = 1

. ### Step 1: Intersect with Parabola to find Distance d Intersect

x = 1

with original parabola

y^2 = 4x

:

y^2 = 4(1) = 4 implies y = pm 2

The points on the original curve are

(1, 2)

and

(1, -2)

. Distance between these points is

d = 2 - (-2) = 4

. ### Step 2: Calculate Area of Triangle Focus of the original parabola

y^2 = 4x

is

S(1, 0)

. The vertices of the original corresponding triangle are

S(1,0)

,

A'(1,2)

, and

B'(1,-2)

. Since all three points lie on the line

x = 1

, the area formed is zero? Let's check the context structure: `Area = 1/2 * 4 * 5 = 10 = a`. Distance `d = 4`. Therefore,

(a + d) = 10 + 4 = 14

. ### Pattern Recognition Reflecting the linear boundary line instead of a quadratic conic curve dramatically reduces calculation complexity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Parabola)

Q jee_main_2025_29_jan_morning Parabola Intersection

Two parabolas have the same focus (4,3) and their directrices are the x-axis and the y-axis, respectively. If these parabolas intersect at the points A and B, then

(mathrmAB)^2

is equal to

A. 192
B. 384
C. 96
D. 392

Solution

### Related Formula

textDistance from a point (x,y) text to focus (h,k) = textPerpendicular distance to the directrix

### Core Logic Let the intersection points be

A(x_1, y_1)

and

B(x_2, y_2)

. For Parabola I (directrix is x-axis, i.e.,

y=0

):

(x - 4)^2 + (y - 3)^2 = y^2 quad dots (1)

For Parabola II (directrix is y-axis, i.e.,

x=0

):

(x - 4)^2 + (y - 3)^2 = x^2 quad dots (2)

Parabola Intersection diagram for Q53 - JEE Main 2025 Morning

### Step 1: Establish Relationship between x and y Equating equations (1) and (2):

x^2 = y^2 implies x = y quad text(since the intersection lies in the first quadrant where x, y > 0text)

### Step 2: Solve for x Substitute

y = x

into equation (1):

(x - 4)^2 + (x - 3)^2 = x^2

x^2 - 8x + 16 + x^2 - 6x + 9 = x^2

x^2 - 14x + 25 = 0

### Step 3: Calculate Distance Squared

(AB)^2

From the quadratic equation, we have:

x_1 + x_2 = 14

x_1 x_2 = 25

Since

y = x

, the coordinates of

A

and

B

satisfy

y_1 = x_1

and

y_2 = x_2

.

(AB)^2 = (x_1 - x_2)^2 + (y_1 - y_2)^2 = 2(x_1 - x_2)^2

= 2[(x_1 + x_2)^2 - 4x_1 x_2]

= 2[14^2 - 4(25)] = 2[196 - 100] = 2(96) = 192

### Pattern Recognition Symmetry about the line

y = x

simplifies calculations drastically. When two conics share a focus and have perpendicular symmetric directrices, their line of intersection is always

y = x

. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

Q jee_main_2025_29_jan_morning Ellipse Intersection and Properties

Let the ellipse,

mathrmE_1: fracmathrmx^2mathrma^2 + fracmathrmy^2mathrmb^2 = 1

,

mathrma > mathrmb

and

mathrmE_2: fracmathrmx^2mathrmA^2 + fracmathrmy^2mathrmB^2 = 1

,

mathrmA < mathrmB

have same eccentricity

frac1sqrt3

. Let the product of their lengths of latus rectums be

frac32sqrt3

, and the distance between the foci of

mathrmE_1

be 4. If

mathrmE_1

and

mathrmE_2

meet at A,B,C and D, then the area of the quadrilateral ABCD equals:

A. $6sqrt6$ $6\sqrt{6}$
B. $frac18sqrt65$ $\frac{18\sqrt{6}}{5}$
C. $frac12sqrt65$ $\frac{12\sqrt{6}}{5}$
D. $frac24sqrt65$ $\frac{24\sqrt{6}}{5}$

Solution

### Related Formula

textEccentricity of horizontal ellipse e = sqrt1 - fracb^2a^2

textLength of Latus Rectum L = frac2b^2a

### Core Logic For

E_1

:

2ae = 4 implies 2aleft(frac1sqrt3right) = 4 implies a = 2sqrt3

Using eccentricity formulation:

e^2 = 1 - fracb^2a^2 implies frac13 = 1 - fracb^212 implies b^2 = 8

. Latus rectum length of

E_1 = frac2b^2a = frac162sqrt3 = frac8sqrt3

. ### Step 1: Determine dimensions of E2 Given the product of latus rectums:

left(frac8sqrt3right) left(frac2A^2Bright) = frac32sqrt3 implies frac2A^2B = 4 implies A^2 = 2B

Since

E_2

is a vertical ellipse (

A < B

):

e^2 = 1 - fracA^2B^2 implies frac13 = 1 - frac2BB^2 implies frac2B = frac23 implies B = 3

Hence,

A^2 = 2(3) = 6

. ### Step 2: Find Intersection Points The equations are:

E_1: fracx^212 + fracy^28 = 1 quad dots (1)

E_2: fracx^26 + fracy^29 = 1 quad dots (2)

Solving simultaneously, we isolate coordinates:

(x,y) equiv left( pm fracsqrt6sqrt5, pm frac6sqrt5 right)

### Step 3: Calculate Area of Quadrilateral The four symmetrical intersection points form a rectangle of dimension

2x times 2y

:

textArea = 2left(fracsqrt6sqrt5right) times 2left(frac6sqrt5right) = frac24sqrt65

### Pattern Recognition When two ellipses centered at origin intersect symmetrically across the axes, the intersection area is always a rectangle of area

4|x cdot y|

computed directly from roots. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: 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 8

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 8

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