Let P(alpha, beta)$P(\alpha, \beta)$ be a point on the parabola y^2 = 4x$y^2 = 4x$. If P$P$ also lies on the chord of the parabola x^2 = 8y$x^2 = 8y$ whose mid point is left(1, frac54right)$\left(1, \frac{5}{4}\right)$, then (alpha - 28)(beta - 8)$(\alpha - 28)(\beta - 8)$ is equal to
Numerical Answer Type:
Enter a numerical valueAnswer: 192 to 192+4 marks
Solution & Explanation
### Related Formula
Equation of chord with given midpoint (x_1, y_1)$(x_1, y_1)$ is T = S_1$T = S_1$.
### Core Logic
For the parabola x^2 = 8y$x^2 = 8y$, the equation of the chord with midpoint left(1, frac54right)$\left(1, \frac{5}{4}\right)$ is:
x(1) - 4left(y + frac54right) = 1^2 - 8left(frac54right)$x(1) - 4\left(y + \frac{5}{4}\right) = 1^2 - 8\left(\frac{5}{4}\right)$x - 4y - 5 = 1 - 10 = -9$x - 4y - 5 = 1 - 10 = -9$x - 4y + 4 = 0 quad dots (i)$x - 4y + 4 = 0 \quad \dots (i)$
### Step 1: Point Intersection with Paraboloid Curve
Since P(alpha, beta)$P(\alpha, \beta)$ lies on this chord and also on y^2 = 4x$y^2 = 4x$:
1) alpha - 4beta + 4 = 0 implies alpha = 4beta - 4$\alpha - 4\beta + 4 = 0 \implies \alpha = 4\beta - 4$
2) beta^2 = 4alpha$\beta^2 = 4\alpha$
Substituting alpha$\alpha$ into the equation:
beta^2 = 4(4beta - 4) implies beta^2 - 16beta + 16 = 0$\beta^2 = 4(4\beta - 4) \implies \beta^2 - 16\beta + 16 = 0$
### Step 2: Calculating the Target Value
We need to find the value of (alpha - 28)(beta - 8)$(\alpha - 28)(\beta - 8)$.
Substitute alpha = 4beta - 4$\alpha = 4\beta - 4$ into this targeted expression:
textValue = (4beta - 4 - 28)(beta - 8) = (4beta - 32)(beta - 8) = 4(beta - 8)(beta - 8) = 4(beta^2 - 16beta + 64)$\text{Value} = (4\beta - 4 - 28)(\beta - 8) = (4\beta - 32)(\beta - 8) = 4(\beta - 8)(\beta - 8) = 4(\beta^2 - 16\beta + 64)$
From the quadratic step, we know beta^2 - 16beta = -16$\beta^2 - 16\beta = -16$. Substituting this:
textValue = 4(-16 + 64) = 4(48) = 192$\text{Value} = 4(-16 + 64) = 4(48) = 192$
### Pattern Recognition
Do not solve for ugly root combinations explicitly if the target expression can be algebraically mapped back to the defining quadratic equations. This prevents unnecessary fractional math steps.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Mathematics: Conic Sections
Keywords:#Chord Midpoint Rule#Conic Section Intersections#Algebraic Substitutions#JEE Main 2024 Coordinate Geometry
More Conic Sections Previous-Year Questions — Page 11
Q23jee_main_2024_30_jan_morningHyperbola
Let the latus rectum of the hyperbolafracx^29 - fracy^2b^2 = 1$\frac{x^2}{9} - \frac{y^2}{b^2} = 1$ subtend an angle of fracpi3$\frac{\pi}{3}$ at the centre of the hyperbola. If b^2$b^2$ is equal to fraclm(1 + sqrtn)$\frac{l}{m}(1 + \sqrt{n})$, where l$l$ and m$m$ are co-prime numbers, then l^2 + m^2 + n^2$l^2 + m^2 + n^2$ is equal to
Numerical Answer.Answer: 182 to 182
Solution
### Related Formula
Latus Rectum = frac2b^2a$Latus Rectum = \frac{2b^2}{a}$e^2 = 1 + fracb^2a^2$e^2 = 1 + \frac{b^2}{a^2}$
### Core Logic
Hyperbola diagram for Q23 - JEE Main 2024 Morning
The endpoints of the latus rectum are (ae, fracb^2a)$(ae, \frac{b^2}{a})$ and (ae, -fracb^2a)$(ae, -\frac{b^2}{a})$.
It subtends 60^circ$60^{\circ}$ at the center (0,0)$(0,0)$.
By symmetry, the line connecting the center to (ae, fracb^2a)$(ae, \frac{b^2}{a})$ makes an angle of 30^circ$30^{\circ}$ with the x-axis.
tan 30^circ = fracfracb^2aae = fracb^2a^2e = frac1sqrt3$\tan 30^{\circ} = \frac{\frac{b^2}{a}}{ae} = \frac{b^2}{a^2e} = \frac{1}{\sqrt{3}}$
### Step 1: Expressing e in terms of b
Given a^2 = 9$a^2 = 9$, so a = 3$a = 3$.
fracb^29e = frac1sqrt3 Rightarrow e = fracsqrt3b^29$\frac{b^2}{9e} = \frac{1}{\sqrt{3}} \Rightarrow e = \frac{\sqrt{3}b^2}{9}$
### Step 2: Applying eccentricity relation
We know e^2 = 1 + fracb^29$e^2 = 1 + \frac{b^2}{9}$.
Substitute e$e$:
left(fracsqrt3b^29right)^2 = 1 + fracb^29$\left(\frac{\sqrt{3}b^2}{9}\right)^2 = 1 + \frac{b^2}{9}$frac3b^481 = 1 + fracb^29$\frac{3b^4}{81} = 1 + \frac{b^2}{9}$fracb^427 = frac9 + b^29$\frac{b^4}{27} = \frac{9 + b^2}{9}$b^4 = 3(9 + b^2) = 27 + 3b^2$b^4 = 3(9 + b^2) = 27 + 3b^2$b^4 - 3b^2 - 27 = 0$b^4 - 3b^2 - 27 = 0$
### Step 3: Solving for b^2
Using the quadratic formula for b^2$b^2$:
b^2 = frac3 pm sqrt9 - 4(1)(-27)2 = frac3 pm sqrt9 + 1082 = frac3 pm sqrt1172$b^2 = \frac{3 \pm \sqrt{9 - 4(1)(-27)}}{2} = \frac{3 \pm \sqrt{9 + 108}}{2} = \frac{3 \pm \sqrt{117}}{2}$
Since b^2 > 0$b^2 > 0$:
b^2 = frac3 + 3sqrt132 = frac32(1 + sqrt13)$b^2 = \frac{3 + 3\sqrt{13}}{2} = \frac{3}{2}(1 + \sqrt{13})$
Comparing this to fraclm(1 + sqrtn)$\frac{l}{m}(1 + \sqrt{n})$:
l = 3$l = 3$, m = 2$m = 2$, n = 13$n = 13$.
### Step 4: Final calculation
Target: l^2 + m^2 + n^2$l^2 + m^2 + n^2$= 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = 182$= 3^2 + 2^2 + 13^2 = 9 + 4 + 169 = 182$
### 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)$(y/x)$.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Conic Sections
Q7jee_main_2024_31_jan_eveningParabola and Ellipse intersection
Let P$P$ be a parabola with vertex (2, 3)$(2, 3)$ and directrix 2x + y = 6$2x + y = 6$. Let an ellipse E:fracx^2a^2 +fracy^2b^2 = 1,a > b$E:\frac{x^2}{a^2} +\frac{y^2}{b^2} = 1,a > b$ of eccentricity frac1sqrt2$\frac{1}{\sqrt{2}}$ pass through the focus of the parabola P$P$. Then the square of the length of the latus rectum of E$E$, 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$\text{Length of Latus Rectum of Ellipse } = \frac{2b^2}{a}$e = sqrt1 - fracb^2a^2$e = \sqrt{1 - \frac{b^2}{a^2}}$
### Core Logic
Parabola and Ellipse intersection diagram for Q7 - JEE Main 2024 Evening
Directrix of parabola is 2x + y - 6 = 0$2x + y - 6 = 0$. Axis is perpendicular to directrix and passes through vertex V(2,3)$V(2,3)$.
Slope of axis = 1/2$= 1/2$. Equation of axis: y - 3 = frac12(x - 2) implies x - 2y + 4 = 0$y - 3 = \frac{1}{2}(x - 2) \implies x - 2y + 4 = 0$.
Intersection of axis and directrix is K$K$:
2x + y = 6$2x + y = 6$x - 2y = -4$x - 2y = -4$
Solving, K = (8/5, 14/5) = (1.6, 2.8)$K = (8/5, 14/5) = (1.6, 2.8)$.
Vertex V(2,3)$V(2,3)$ is the midpoint of focus S(alpha, beta)$S(\alpha, \beta)$ and K(1.6, 2.8)$K(1.6, 2.8)$:
fracalpha + 1.62 = 2 implies alpha = 2.4$\frac{\alpha + 1.6}{2} = 2 \implies \alpha = 2.4$fracbeta + 2.82 = 3 implies beta = 3.2$\frac{\beta + 2.8}{2} = 3 \implies \beta = 3.2$
So, focus S = (2.4, 3.2)$S = (2.4, 3.2)$.
Ellipse passes through S(2.4, 3.2)$S(2.4, 3.2)$:
frac(2.4)^2a^2 + frac(3.2)^2b^2 = 1$\frac{(2.4)^2}{a^2} + \frac{(3.2)^2}{b^2} = 1$
Eccentricity e = frac1sqrt2 implies 1 - fracb^2a^2 = frac12 implies a^2 = 2b^2$e = \frac{1}{\sqrt{2}} \implies 1 - \frac{b^2}{a^2} = \frac{1}{2} \implies a^2 = 2b^2$.
Substitute a^2 = 2b^2$a^2 = 2b^2$:
frac5.762b^2 + frac10.24b^2 = 1 implies frac2.88 + 10.24b^2 = 1 implies b^2 = 13.12 = frac32825$\frac{5.76}{2b^2} + \frac{10.24}{b^2} = 1 \implies \frac{2.88 + 10.24}{b^2} = 1 \implies b^2 = 13.12 = \frac{328}{25}$
Square of Latus Rectum:
L^2 = left(frac2b^2aright)^2 = frac4b^4a^2 = frac4b^42b^2 = 2b^2 = 2 times frac32825 = frac65625$L^2 = \left(\frac{2b^2}{a}\right)^2 = \frac{4b^4}{a^2} = \frac{4b^4}{2b^2} = 2b^2 = 2 \times \frac{328}{25} = \frac{656}{25}$
### 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
Q3jee_main_2024_31_jan_morningHyperbola and Ellipse Properties
If the foci of a hyperbola are same as that of the ellipse fracx^29 + fracy^225 = 1$\frac{x^2}{9} + \frac{y^2}{25} = 1$ and the eccentricity of the hyperbola is frac158$\frac{15}{8}$ times the eccentricity of the ellipse, then the smaller focal distance of the point left(sqrt2, frac143sqrtfrac25right)$\left(\sqrt{2}, \frac{14}{3}\sqrt{\frac{2}{5}}\right)$ on the hyperbola, is equal to
### Related Formula
e = sqrt1 - fraca^2b^2 text (for vertical ellipse)$e = \sqrt{1 - \frac{a^2}{b^2}} \text{ (for vertical ellipse)}$textFocal distance of P(x_1, y_1) text on hyperbola = e_H |y_1 pm fracBe_H|$\text{Focal distance of } P(x_1, y_1) \text{ on hyperbola} = e_H |y_1 \pm \frac{B}{e_H}|$
### Core Logic
For the ellipse fracx^29 + fracy^225 = 1$\frac{x^2}{9} + \frac{y^2}{25} = 1$:
a = 3, b = 5$a = 3, b = 5$. Since b > a$b > a$, the major axis is along the y-axis.
e = sqrt1 - frac925 = frac45$e = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$
Foci = (0, pm be) = (0, pm 4)$= (0, \pm be) = (0, \pm 4)$.
### Step 1: Hyperbola Properties
Eccentricity of hyperbolae_H = frac45 times frac158 = frac32$e_H = \frac{4}{5} \times \frac{15}{8} = \frac{3}{2}$.
Let the hyperbola be fracx^2A^2 - fracy^2B^2 = -1$\frac{x^2}{A^2} - \frac{y^2}{B^2} = -1$ since its foci (0, pm 4)$(0, \pm 4)$ are on the y-axis.
Foci B e_H = 4 implies B = frac83$B e_H = 4 \implies B = \frac{8}{3}$.
A^2 = B^2(e_H^2 - 1) = frac649left(frac94 - 1right) = frac809$A^2 = B^2(e_H^2 - 1) = \frac{64}{9}\left(\frac{9}{4} - 1\right) = \frac{80}{9}$.
Equation: fracx^280/9 - fracy^264/9 = -1$\frac{x^2}{80/9} - \frac{y^2}{64/9} = -1$.
### Step 2: Focal Distance Calculation
Directrix of hyperbola: y = pm fracBe_H = pm frac169$y = \pm \frac{B}{e_H} = \pm \frac{16}{9}$.
Smaller focal distance PS = e_H cdot PM$PS = e_H \cdot PM$, where PM$PM$ is the distance to the nearer directrix.
PS = frac32 left| frac143sqrtfrac25 - frac169 right| = 7sqrtfrac25 - frac83$PS = \frac{3}{2} \left| \frac{14}{3}\sqrt{\frac{2}{5}} - \frac{16}{9} \right| = 7\sqrt{\frac{2}{5}} - \frac{8}{3}$
### Pattern Recognition
Smaller focal distance of a point (x_1, y_1)$(x_1, y_1)$ on a vertical hyperbola is given by e|y_1 - fracbe|$e|y_1 - \frac{b}{e}|$ where y_1 > 0$y_1 > 0$ and we use the positive directrix.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 11 Maths: Conic Sections
Q27jee_main_2024_31_jan_morningEllipse and Hyperbola Properties
Let the foci and length of the latus rectum of an ellipse fracx^2a^2 + fracy^2b^2 = 1, a > b$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1, a > b$ be (pm 5, 0)$(\pm 5, 0)$ and sqrt50$\sqrt{50}$, respectively. Then, the square of the eccentricity of the hyperbola fracx^2b^2 - fracy^2a^2 b^2 = 1$\frac{x^2}{b^2} - \frac{y^2}{a^2 b^2} = 1$ equals
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