Solution & Explanation
### Related Formula
Centroid (h, k)$(h, k)$ of a triangle with vertices (x_1,y_1), (x_2,y_2), (x_3,y_3)$(x_1,y_1), (x_2,y_2), (x_3,y_3)$:
h = fracx_1 + x_2 + x_33, quad k = fracy_1 + y_2 + y_33$h = \frac{x_1 + x_2 + x_3}{3}, \quad k = \frac{y_1 + y_2 + y_3}{3}$
### Core Logic
Given equations:
1) Circle: y^2 = 8x - x^2$y^2 = 8x - x^2$
2) Hyperbola: 4x^2 - 9y^2 = 36$4x^2 - 9y^2 = 36$
Substitute circle's y^2$y^2$ into hyperbola equation:
4x^2 - 9(8x - x^2) = 36 implies 4x^2 - 72x + 9x^2 = 36$4x^2 - 9(8x - x^2) = 36 \implies 4x^2 - 72x + 9x^2 = 36$
13x^2 - 72x - 36 = 0 implies (13x + 6)(x - 6) = 0$13x^2 - 72x - 36 = 0 \implies (13x + 6)(x - 6) = 0$
If x = -6/13$x = -6/13$, y^2 < 0$y^2 < 0$ (rejected).
Thus, x = 6$x = 6$.
Substituting x = 6$x = 6$ into circle: y^2 = 8(6) - 6^2 = 48 - 36 = 12 implies y = pm sqrt12$y^2 = 8(6) - 6^2 = 48 - 36 = 12 \implies y = \pm \sqrt{12}$.
The intersection points are A(6, sqrt12)$A(6, \sqrt{12})$ and B(6, -sqrt12)$B(6, -\sqrt{12})$.
### Step 1: Relate Centroid coordinates to P
Let point P$P$ have coordinates (alpha, beta)$(\alpha, \beta)$. Since P$P$ lies on 2x - 3y + 4 = 0$2x - 3y + 4 = 0$:
2alpha - 3beta + 4 = 0 implies beta = frac2alpha + 43$2\alpha - 3\beta + 4 = 0 \implies \beta = \frac{2\alpha + 4}{3}$
Let the centroid be (h, k)$(h, k)$:
h = frac6 + 6 + alpha3 = frac12 + alpha3 implies alpha = 3h - 12$h = \frac{6 + 6 + \alpha}{3} = \frac{12 + \alpha}{3} \implies \alpha = 3h - 12$
k = fracsqrt12 - sqrt12 + beta3 = fracbeta3 implies beta = 3k$k = \frac{\sqrt{12} - \sqrt{12} + \beta}{3} = \frac{\beta}{3} \implies \beta = 3k$
### Step 2: Form the Locus Equation
Substitute alpha$\alpha$ and \beta into the line equation of P$P$:
2(3h - 12) - 3(3k) + 4 = 0$2(3h - 12) - 3(3k) + 4 = 0$
6h - 24 - 9k + 4 = 0$6h - 24 - 9k + 4 = 0$
6h - 9k = 20$6h - 9k = 20$
Replacing (h, k)$(h, k)$ with general coordinates (x, y)$(x, y)$ gives the locus:
6x - 9y = 20$6x - 9y = 20$
### Pattern Recognition
Notice how the y$y$-coordinates of intersection points A$A$ and B$B$ are symmetric (\,pmsqrt12\,$\,\pm\sqrt{12}\,$), meaning their sum is zero. This simplifies the expression for k$k$ instantly to just beta/3$\beta/3$.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Coordinate Geometry
Class 11 Mathematics: 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$x^2 + (y - 1)^2 = 2$, E_1$E_1$ and E_2$E_2$ 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$x + y = 3$ touch the curves C, E_1$E_1$ and E_2$E_2$ at P(x_1, y_1)$P(x_1, y_1)$, Q(x_2, y_2)$Q(x_2, y_2)$ and R(x_3, y_3)$R(x_3, y_3)$ respectively. Given that P is the mid-point of the line segment QR and PQ = frac2sqrt23$PQ = \frac{2\sqrt{2}}{3}$, the value of 9(x_1y_1 + x_2y_2 + x_3y_3)$9(x_1y_1 + x_2y_2 + x_3y_3)$ 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$x = x_1 + r\cos\theta, \quad y = y_1 + r\sin\theta$
### Core Logic
Step 1: Find point P(x_1,y_1)$P(x_1,y_1)$ on circle C$C$.
Equation of tangent at P$P$ on x^2 + y^2 - 2y - 1 = 0$x^2 + y^2 - 2y - 1 = 0$ is xx_1 + y(y_1 - 1) - (y_1 + 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$x + y = 3 \implies \frac{x_1}{1} = \frac{y_1 - 1}{1} = \frac{y_1 + 1}{3}$.
Solving gives x_1 = 1, y_1 = 2$x_1 = 1, y_1 = 2$. Thus, P = (1, 2)$P = (1, 2)$.
### Step 1: Use Line Parametrics for Q and R
Line x + y = 3$x + y = 3$ makes an angle theta = 135^circ$\theta = 135^{\circ}$ with the positive x-axis.
Using parametric distances from P(1,2)$P(1,2)$ with r = PQ = frac2sqrt23$r = PQ = \frac{2\sqrt{2}}{3}$:
x = 1 pm rcos(135^circ) = 1 mp fracrsqrt2$x = 1 \pm r\cos(135^{\circ}) = 1 \mp \frac{r}{\sqrt{2}}$
y = 2 pm rsin(135^circ) = 2 pm fracrsqrt2$y = 2 \pm r\sin(135^{\circ}) = 2 \pm \frac{r}{\sqrt{2}}$
Substitute r = frac2sqrt23$r = \frac{2\sqrt{2}}{3}$:
For Q$Q$: x_2 = 1 + frac23 = frac53, y_2 = 2 - frac23 = frac43$x_2 = 1 + \frac{2}{3} = \frac{5}{3}, y_2 = 2 - \frac{2}{3} = \frac{4}{3}$.
For R$R$: x_3 = 1 - frac23 = frac13, y_3 = 2 + frac23 = frac83$x_3 = 1 - \frac{2}{3} = \frac{1}{3}, y_3 = 2 + \frac{2}{3} = \frac{8}{3}$.
### Step 2: Evaluate Final Expression
Calculate the products:
x_1y_1 = 1 times 2 = 2$x_1y_1 = 1 \times 2 = 2$
x_2y_2 = frac53 times frac43 = frac209$x_2y_2 = \frac{5}{3} \times \frac{4}{3} = \frac{20}{9}$
x_3y_3 = frac13 times frac83 = frac89$x_3y_3 = \frac{1}{3} \times \frac{8}{3} = \frac{8}{9}$
9(x_1y_1 + x_2y_2 + x_3y_3) = 9left(2 + frac209 + frac89right) = 18 + 20 + 8 = 46$9(x_1y_1 + x_2y_2 + x_3y_3) = 9\left(2 + \frac{20}{9} + \frac{8}{9}\right) = 18 + 20 + 8 = 46$
### Pattern Recognition
Parametric distance equations are perfect for lines containing midpoints. This approach bypasses calculating the individual ellipse equations a^2, b^2$a^2, b^2$ completely.
### Evaluation Rubric / Model Answer
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### 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$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be 10. If its eccentricity is the minimum value of the function f(t) = t^2 + t + frac1112, t in mathbfR$f(t) = t^2 + t + \frac{11}{12}, t \in \mathbf{R}$, then a^2 + b^2$a^2 + b^2$ 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$\text{LR} = \frac{2b^2}{a}$
e^2 = 1 - fracb^2a^2$e^2 = 1 - \frac{b^2}{a^2}$
### Core Logic
Given length of textLR = 10 implies frac2b^2a = 10 implies b^2 = 5a quad dots text(i)$\text{LR} = 10 \implies \frac{2b^2}{a} = 10 \implies b^2 = 5a \quad \dots \text{(i)}$
Now, let's find the minimum value of f(t) = t^2 + t + frac1112$f(t) = t^2 + t + \frac{11}{12}$.
Differentiating: f'(t) = 2t + 1 = 0 implies t = -frac12$f'(t) = 2t + 1 = 0 \implies t = -\frac{1}{2}$.
textMinimum value e = fleft(-frac12
ight) = left(-frac12
ight)^2 + left(-frac12
ight) + frac1112 = frac14 - frac12 + frac1112 = frac3 - 6 + 1112 = frac812 = frac23$\text{Minimum value } e = f\left(-\frac{1}{2}
ight) = \left(-\frac{1}{2}
ight)^2 + \left(-\frac{1}{2}
ight) + \frac{11}{12} = \frac{1}{4} - \frac{1}{2} + \frac{11}{12} = \frac{3 - 6 + 11}{12} = \frac{8}{12} = \frac{2}{3}$
### 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)$e^2 = \frac{4}{9} = 1 - \frac{b^2}{a^2} \implies \frac{b^2}{a^2} = \frac{5}{9} \implies b^2 = \frac{5a^2}{9} \quad \dots \text{(ii)}$
Equating (i) and (ii):
5a = frac5a^29 implies a = 9$5a = \frac{5a^2}{9} \implies a = 9$
Then from (i):
b^2 = 5(9) = 45 implies b = 3sqrt5$b^2 = 5(9) = 45 \implies b = 3\sqrt{5}$
Hence, a^2 = 81$a^2 = 81$.
### Step 2: Calculate a^2 + b^2
a^2 + b^2 = 81 + 45 = 126$a^2 + b^2 = 81 + 45 = 126$
### Pattern Recognition
A quadratic function at^2+bt+c$at^2+bt+c$ reaches its extreme value at t = -fracb2a$t = -\frac{b}{2a}$. Using this layout avoids full calculus derivation steps.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 11 Mathematics: Conic Sections
Class 11 Mathematics: Quadratic Equations
Q68
2025
Eccentricity and Foci
Let mathbfe_1$\mathbf{e}_1$ and mathbfe_2$\mathbf{e}_2$ be the eccentricities of the ellipse fracmathrmx^2mathrmb^2 + fracmathrmy^225 = 1$\frac{\mathrm{x}^2}{\mathrm{b}^2} + \frac{\mathrm{y}^2}{25} = 1$ and the hyperbola fracmathrmx^216 - fracmathrmy^2mathrmb^2 = 1$\frac{\mathrm{x}^2}{16} - \frac{\mathrm{y}^2}{\mathrm{b}^2} = 1$, respectively. If mathrmb < 5$\mathrm{b} < 5$ and mathrme_1mathrme_2 = 1$\mathrm{e}_1\mathrm{e}_2 = 1$, 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)$(-5,0)$ and 5x + 9 = 0$5x + 9 = 0$, respectively. If the product of the focal distances of a point (alpha,2sqrt5)$(\alpha,2\sqrt{5})$ on the hyperbola is p$p$, then 4p$4p$ 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$PF_1 \cdot PF_2 = e^2\alpha^2 - a^2$
### Core Logic
Given focus ae = 5$ae = 5$ and directrix fracae = frac95$\frac{a}{e} = \frac{9}{5}$.
Multiplying gives a^2 = 9 implies a = 3$a^2 = 9 \implies a = 3$.
Then 3e = 5 implies e = frac53$3e = 5 \implies e = \frac{5}{3}$.
Using hyperbola identity: b^2 = a^2(e^2 - 1) = 9left(frac259 - 1right) = 16 implies b = 4$b^2 = a^2(e^2 - 1) = 9\left(\frac{25}{9} - 1\right) = 16 \implies b = 4$.
The equation of the hyperbola is:
fracx^29 - fracy^216 = 1$\frac{x^2}{9} - \frac{y^2}{16} = 1$
### Step 1: Point Substitution
Since point (alpha, 2sqrt5)$(\alpha, 2\sqrt{5})$ lies on the hyperbola:
fracalpha^29 - frac2016 = 1 implies fracalpha^29 = 1 + frac54 = frac94 implies alpha^2 = frac814$\frac{\alpha^2}{9} - \frac{20}{16} = 1 \implies \frac{\alpha^2}{9} = 1 + \frac{5}{4} = \frac{9}{4} \implies \alpha^2 = \frac{81}{4}$
### Step 2: Focal Product Calculation
Evaluating p$p$:
p = e^2alpha^2 - a^2 = left(frac259right)left(frac814right) - 9 = frac2254 - 9 = frac1894$p = e^2\alpha^2 - a^2 = \left(\frac{25}{9}\right)\left(\frac{81}{4}\right) - 9 = \frac{225}{4} - 9 = \frac{189}{4}$
4p = 189$4p = 189$
### Pattern Recognition
Combining the metric coordinates ae$ae$ and fracae$\frac{a}{e}$ via simple multiplication locks in the basic structural axis parameter a^2$a^2$ 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$\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, whose mid-point is (3,1)$(3,1)$ 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)$(x_1, y_1)$ is given is:
T = S_1$T = S_1$
fracxx_1a^2 + fracyy_1b^2 = fracx_1^2a^2 + fracy_1^2b^2$\frac{xx_1}{a^2} + \frac{yy_1}{b^2} = \frac{x_1^2}{a^2} + \frac{y_1^2}{b^2}$
### Core Logic
For the given ellipse fracx^225 + fracy^216 = 1$\frac{x^2}{25} + \frac{y^2}{16} = 1$ and midpoint (x_1, y_1) = (3, 1)$(x_1, y_1) = (3, 1)$ , we substitute these values into the T = S_1$T = S_1$ expression.
### Step 1: Substitution and Expansion
Substituting the coordinates into the formula:
frac3x25 + frac1y16 - 1 = frac3^225 + frac1^216 - 1$\frac{3x}{25} + \frac{1y}{16} - 1 = \frac{3^2}{25} + \frac{1^2}{16} - 1$
frac3x25 + fracy16 = frac925 + frac116$\frac{3x}{25} + \frac{y}{16} = \frac{9}{25} + \frac{1}{16}$
### 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)$16(3x) + 25(y) = 16(9) + 25(1)$
48x + 25y = 144 + 25$48x + 25y = 144 + 25$
48x + 25y = 169$48x + 25y = 169$
### Pattern Recognition
Whenever a midpoint is given for a chord of any second-degree conic curve, the relation T = S_1$T = S_1$ simplifies the process instantaneously without finding the individual intersection points.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
