A line passing through the point $P(sqrt5, sqrt5)$ $P(\sqrt{5}, \sqrt{5})$ intersects the ellipse $fracx^236 + fracy^225 = 1$ $\frac{x^2}{36} + \frac{y^2}{25} = 1$ at $A$ $A$ and $B$ $B$ [cite: 567] such that $(PA) cdot (PB)$ $(PA) \cdot (PB)$ is maximum. Then $5(PA^2 + PB^2)$ $5(PA^{2} + PB^{2})$ is equal to[cite: 570]:

A. 218
B. 377
C. 290
D. 338

Solution & Explanation

### Related Formula Parametric line equation relative to an offset point

P(x_0, y_0)

x = x_0 + rcostheta, quad y = y_0 + rsintheta

Ellipse and Line Properties diagram for Q56 - JEE Main 2025 Morning

### Core Logic Assume any line through

P

can be represented parametrically by [cite: 1277]:

Q(sqrt5 + rcostheta, sqrt5 + rsintheta)

[cite: 1277] Substitute coordinates into the standard ellipse formula [cite: 1278]:

25(sqrt5 + rcostheta)^2 + 36(sqrt5 + rsintheta)^2 = 900

[cite: 1278] Expanding and gathering powers of

r

yields [cite: 1280]:

r^2(25cos^2theta + 36sin^2theta) + 2sqrt5r(25costheta + 36sintheta) - 595 = 0

[cite: 1280] The product of roots corresponds to the distance product value[cite: 1281, 1283]:

PA cdot PB = |r_1 r_2| = frac59525cos^2theta + 36sin^2theta = frac59525 + 11sin^2theta

[cite: 1283] ### Step 1: Maximization condition To make

PA cdot PB

maximum, the denominator must be minimized [cite: 1284]:

sin^2theta = 0 implies theta = 0

[cite: 1284] This implies the chord line

AB

must run parallel to the x-axis [cite: 1285]:

y_A = y_B = sqrt5

[cite: 1285] Substitute

y = sqrt5

back into the ellipse equation to calculate x-coordinates [cite: 1286]:

fracx^236 + frac525 = 1 implies fracx^236 = frac45 implies x^2 = frac1445

[cite: 1287] Therefore, the coordinates are

x = pm frac12sqrt5

. ### Step 2: Distance value summation Compute

PA^2 + PB^2

using coordinates directly [cite: 1289]:

PA^2 + PB^2 = left(sqrt5 - frac12sqrt5right)^2 + left(sqrt5 + frac12sqrt5right)^2

[cite: 1289]

= 2left(5 + frac1445right) = frac3385

[cite: 1290] Multiplying by 5 gives the target integer answer [cite: 1290]:

5(PA^2 + PB^2) = 338

[cite: 1290] ### Pattern Recognition Parametric distances from a point intersecting a conic configuration usually form a standard quadratic equation in

r

. The angle parameter immediately simplifies the boundary constraint optimization. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Mathematics: Conic Sections (Ellipse)

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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