The centre of a circle $C$ $C$ is at the centre of the ellipse $E: fracx^2a^2 + fracy^2b^2 = 1, a > b$ . Let $C$ $C$ pass through the foci $F_1$ $F_1$ and $F_2$ $F_2$ of $E such that the circle$ $E such that the circle $C $and the ellipse$ $ and the ellipse $E $intersect at four points. Let$ $ intersect at four points. Let $P $be one of these four points. If the area of the triangle$ PF_1F_2 $is 30 and the length of the major axis of$ E $is 17, then the distance between the foci of$ E$ is :

A. $26$ $26$
B. $13$ $13$
C. $12$ $12$
D. $frac132$ $\frac{13}{2}$

Solution & Explanation

### Core Logic The circle

C

has its center at the origin and passes through the foci

F_1(-ae, 0)

and

F_2(ae, 0)

. This means

F_1F_2

is the diameter of the circle. Any point

P

lying on this circle satisfies the property that the angle subtended by the diameter is a right angle:

angle F_1PF_2 = 90^circ

Ellipse properties diagram for Q67 - JEE Main 2025 Evening

### Step 1: Using the Area and Ellipse Definition Since

triangle PF_1F_2

is a right-angled triangle at

P

textArea = frac12 cdot PF_1 cdot PF_2 = 30 implies PF_1 cdot PF_2 = 60

By the definition of an ellipse, the sum of the focal distances to any point on the curve is equal to the length of the major axis (

2a = 17

PF_1 + PF_2 = 17

### Step 2: Calculating Distance between Foci Applying Pythagoras' theorem in right-angled

triangle PF_1F_2

F_1F_2^2 = PF_1^2 + PF_2^2 = (PF_1 + PF_2)^2 - 2(PF_1 cdot PF_2)

Substitute the known values from our equations block:

F_1F_2^2 = (17)^2 - 2(60) = 289 - 120 = 169

F_1F_2 = sqrt169 = 13

Therefore, the distance between the foci is

13

. ### Pattern Recognition Whenever a circle is circumscribed around the foci of an ellipse, remember Thales' theorem: any intersection point with the ellipse forms a right triangle with the focal diameter, linking focal properties directly to Pythagoras. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

Reference Study Guides

Conic Sections

More Conic Sections Previous-Year Questions — Page 6

Q74 2025 Tangents to Conics

Let C be the circle

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

E_1

and

E_2

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,

E_1

and

E_2

P(x_1, y_1)

Q(x_2, y_2)

and

R(x_3, y_3)

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

PQ = frac2sqrt23

, the value of

9(x_1y_1 + x_2y_2 + x_3y_3)

is equal to

Numerical Answer. Answer: 46 to 46

Solution

### Related Formula Parametric equation of a straight line:

x = x_1 + rcostheta, quad y = y_1 + rsintheta

### Core Logic Step 1: Find point

P(x_1,y_1)

on circle

C

. Equation of tangent at

P

x^2 + y^2 - 2y - 1 = 0

xx_1 + y(y_1 - 1) - (y_1 + 1) = 0

. Comparing with line

x + y = 3 implies fracx_11 = fracy_1 - 11 = fracy_1 + 13

. Solving gives

x_1 = 1, y_1 = 2

. Thus,

P = (1, 2)

. ### Step 1: Use Line Parametrics for Q and R Line

x + y = 3

makes an angle

theta = 135^circ

with the positive x-axis. Using parametric distances from

P(1,2)

with

r = PQ = frac2sqrt23

x = 1 pm rcos(135^circ) = 1 mp fracrsqrt2

y = 2 pm rsin(135^circ) = 2 pm fracrsqrt2

Substitute

r = frac2sqrt23

: For

Q

x_2 = 1 + frac23 = frac53, y_2 = 2 - frac23 = frac43

. For

R

x_3 = 1 - frac23 = frac13, y_3 = 2 + frac23 = frac83

. ### Step 2: Evaluate Final Expression Calculate the products:

x_1y_1 = 1 times 2 = 2

x_2y_2 = frac53 times frac43 = frac209

x_3y_3 = frac13 times frac83 = frac89

9(x_1y_1 + x_2y_2 + x_3y_3) = 9left(2 + frac209 + frac89right) = 18 + 20 + 8 = 46

### Pattern Recognition Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations

a^2, b^2

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

Q62 2025 Properties of Ellipse

Let the length of a latus rectum of an ellipse

fracx^2a^2 + fracy^2b^2 = 1

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

f(t) = t^2 + t + frac1112, t in mathbfR

, then

a^2 + b^2

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:

textLR = frac2b^2a

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

### Core Logic Given length of

textLR = 10 implies frac2b^2a = 10 implies b^2 = 5a quad dots text(i)

Now, let's find the minimum value of

f(t) = t^2 + t + frac1112

. Differentiating:

f'(t) = 2t + 1 = 0 implies t = -frac12

textMinimum value e = fleft(-frac12 ight) = left(-frac12 ight)^2 + left(-frac12 ight) + frac1112 = frac14 - frac12 + frac1112 = frac3 - 6 + 1112 = frac812 = frac23

### Step 1: Solve for a and b Using eccentricity formula:

e^2 = frac49 = 1 - fracb^2a^2 implies fracb^2a^2 = frac59 implies b^2 = frac5a^29 quad dots text(ii)

Equating (i) and (ii):

5a = frac5a^29 implies a = 9

Then from (i):

b^2 = 5(9) = 45 implies b = 3sqrt5

Hence,

a^2 = 81

. ### Step 2: Calculate a^2 + b^2

a^2 + b^2 = 81 + 45 = 126

### Pattern Recognition A quadratic function

at^2+bt+c

reaches its extreme value at

t = -fracb2a

. Using this layout avoids full calculus derivation steps. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections Class 11 Mathematics: Quadratic Equations

Q68 2025 Eccentricity and Foci

Let

mathbfe_1

and

mathbfe_2

be the eccentricities of the ellipse

fracmathrmx^2mathrmb^2 + fracmathrmy^225 = 1

and the hyperbola

fracmathrmx^216 - fracmathrmy^2mathrmb^2 = 1

, respectively. If

mathrmb < 5

and

mathrme_1mathrme_2 = 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. $frac45$ $\frac{4}{5}$
B. $frac35$ $\frac{3}{5}$
C. $fracsqrt74$ $\frac{\sqrt{7}}{4}$
D. $fracsqrt32$ $\frac{\sqrt{3}}{2}$

Solution

### Related Formula Eccentricity for ellipse (

a$a) and hyperbola are calculated as: e_1^2 = 1 - fraca^2b^2$e_1^2 = 1 - \frac{a^2}{b^2}$ e_2^2 = 1 + fracb^2a^2$e_2^2 = 1 + \frac{b^2}{a^2}$ ### Core Logic Since b < 5$b < 5$, for the ellipse fracx^2b^2 + fracy^225 = 1$\frac{x^2}{b^2} + \frac{y^2}{25} = 1$, the major axis is along the y$y$-axis: e_1^2 = 1 - fracb^225$e_1^2 = 1 - \frac{b^2}{25}$ For the hyperbola fracx^216 - fracy^2b^2 = 1$\frac{x^2}{16} - \frac{y^2}{b^2} = 1$: e_2^2 = 1 + fracb^216$e_2^2 = 1 + \frac{b^2}{16}$ Given e_1^2 e_2^2 = 1$e_1^2 e_2^2 = 1$: left(1 - fracb^225right)left(1 + fracb^216right) = 1$\left(1 - \frac{b^2}{25}\right)\left(1 + \frac{b^2}{16}\right) = 1$ 1 + fracb^216 - fracb^225 - fracb^4400 = 1 implies frac9b^2400 = fracb^4400 implies b^2 = 9$1 + \frac{b^2}{16} - \frac{b^2}{25} - \frac{b^4}{400} = 1 \implies \frac{9b^2}{400} = \frac{b^4}{400} \implies b^2 = 9$ ### Step 1: Calculate Foci Locations Substituting b^2 = 9$b^2 = 9$: Ellipse foci: ae_1 = 5 cdot sqrt1 - frac925 = 5 cdot frac45 = 4$ae_1 = 5 \cdot \sqrt{1 - \frac{9}{25}} = 5 \cdot \frac{4}{5} = 4$. Foci lie along y$y$-axis: (0, pm 4)$(0, \pm 4)$. Hyperbola foci: ae_2 = 4 cdot sqrt1 + frac916 = 4 cdot frac54 = 5$ae_2 = 4 \cdot \sqrt{1 + \frac{9}{16}} = 4 \cdot \frac{5}{4} = 5$. 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 = sqrt1 - fracB^2A^2 = sqrt1 - frac1625 = frac35$E = \sqrt{1 - \frac{B^2}{A^2}} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5}$ ### Pattern Recognition When a conic passes through points directly on the axes like (pm alpha, 0)$(\pm alpha, 0)$ and (0, pm beta)$(0, \pm \beta)$, those points immediately represent the semi-axes values A$A$ and B$B$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

Q74 2025 Properties of Hyperbola

Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be

(-5,0)

and

5x + 9 = 0

, respectively. If the product of the focal distances of a point

(alpha,2sqrt5)

on the hyperbola is

p

, then

4p

is equal to

Numerical Answer. Answer: 189 to 189

Solution

### Related Formula Product of focal distances for a point on a hyperbola satisfies:

PF_1 cdot PF_2 = e^2alpha^2 - a^2

### Core Logic Given focus

ae = 5

and directrix

fracae = frac95

. Multiplying gives

a^2 = 9 implies a = 3

. Then

3e = 5 implies e = frac53

. Using hyperbola identity:

b^2 = a^2(e^2 - 1) = 9left(frac259 - 1right) = 16 implies b = 4

. The equation of the hyperbola is:

fracx^29 - fracy^216 = 1

### Step 1: Point Substitution Since point

(alpha, 2sqrt5)

lies on the hyperbola:

fracalpha^29 - frac2016 = 1 implies fracalpha^29 = 1 + frac54 = frac94 implies alpha^2 = frac814

### Step 2: Focal Product Calculation Evaluating

p

p = e^2alpha^2 - a^2 = left(frac259right)left(frac814right) - 9 = frac2254 - 9 = frac1894

4p = 189

### Pattern Recognition Combining the metric coordinates

ae

and

fracae

via simple multiplication locks in the basic structural axis parameter

a^2

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

Q51 2025 Ellipse - Equation of Chord with Given Midpoint

The equation of the chord, of the ellipse

fracx^225+fracy^216=1

, whose mid-point is

(3,1)

is: [cite: 3245, 3246]

A. $48x+25y=169$ $48x+25y=169$
B. $4x+122y=134$ $4x+122y=134$
C. $25x+101y=176$ $25x+101y=176$
D. $5x+16y=31$ $5x+16y=31$

Solution

### Related Formula The equation of a chord of an ellipse whose midpoint

(x_1, y_1)

is given is:

T = S_1

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

### Core Logic For the given ellipse

fracx^225 + fracy^216 = 1

and midpoint

(x_1, y_1) = (3, 1)

, we substitute these values into the

T = S_1

expression. ### Step 1: Substitution and Expansion Substituting the coordinates into the formula:

frac3x25 + frac1y16 - 1 = frac3^225 + frac1^216 - 1

frac3x25 + fracy16 = frac925 + frac116

### Step 2: Simplification Multiply both sides by the least common multiple of 25 and 16, which is 400:

16(3x) + 25(y) = 16(9) + 25(1)

48x + 25y = 144 + 25

48x + 25y = 169

### Pattern Recognition Whenever a midpoint is given for a chord of any second-degree conic curve, the relation

T = S_1

simplifies the process instantaneously without finding the individual intersection points. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections

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