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Maths Class 11th

A parabola has vertex (2, 3), equation of directrix is 2x – y = 1 and equation of ellipse is $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, e = \frac{1}{\sqrt[]{2}}$ and ellipse passing through focur of parabola then square of length of latus rectum of ellipse is

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alok kumar singh

Contributor-Level 10

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

...more

Slope of axis = $\frac{1}{2}$

$y - 3 = \frac{1}{2} (x - 2)$

⇒ 2y – 6 = x – 2

⇒ 2y – x – 4 = 0

2x + y – 6 = 0

4x + 2y – 12 = 0

α + 1.6 = 4 ⇒ α = 2.4

β + 2.8 = 6 ⇒ β = 3.2

Ellipse passes through (2.4, 3.2)

⇒ $\frac{{(\frac{24}{10})}^{2}}{a^{2}} + \frac{{(\frac{32}{10})}^{2}}{b^{2}} = 1$

Also $1 - \frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{a^{2}}{b^{2}} = \frac{1}{2}$

$\frac{144}{25} b^{2} + \frac{256}{25} a^{2} = a^{2} b^{2}$

$\frac{144}{25} + \frac{256}{25} * \frac{1}{2} = a^{2}$

$\frac{(128 + 144)}{25} = a^{2} \Rightarrow \frac{272}{25} = a^{2}$

-> $b^{2} = \frac{2 * 272}{25}$

Latus rectum = $\frac{2 b^{2}}{a}$

(Latus rectum)²

$= \frac{4 b^{4}}{a^{2}} = 4 (\frac{b^{2}}{a^{2}}) b^{2} = \frac{8 * 272 * 2}{25} = \frac{4352}{25}$

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New answer posted

a month ago

Maths Class 12th

If (α, β, γ) is mirror image of the point (2, 3, 4) with respect to the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ Then 2α + 3β + 4γ is

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alok kumar singh

Contributor-Level 10

Take $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = λ$

x = 2λ + 1, y = 3λ + 2, z = 4λ + 3

$\vec{A B}$ = (α − 2) $\hat{i}$ + (β − 3) $\hat{j}$ + (γ − 4) $\hat{k}$

Now,

(α − 2) ⋅ 2 + (β − 3) ⋅3 + (γ − 4) ⋅ 4 = 0

2α − 4 + 3β − 9 + 4γ −16 = 0

⇒ 2α + 3β + 4γ = 29

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New answer posted

a month ago

Maths Inverse Trigonometric Functions

Let f : (−∞, −1] → (a, b] be defined as $f (x) = e^{x^{3} - 3 x + 1}$ , if f is both one and onto, then the distance from a point P(2a + 4, b + 2) to curve x + ye⁻³ − 4 = 0 is

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alok kumar singh

Contributor-Level 10

$f (x) = e^{x^{3} - 3 x + 1}$

$f' (x) = e^{x^{3} - 3 x + 1}$ (3x² − 3)

$e^{x^{3} - 3 x + 1}$ = ⋅ 3(x −1)(x +1)

For x ∈ (−∞, −1], f '(x) ≥ 0

∴ f(x) is increasing function

∴ a = e^–∞ = 0 = f (−∞)

b = e⁻¹⁺³⁺¹ = e³ = f (−1)

∴ P(4, e³ + 2)

$d = \frac{(e^{3} + 2) (e^{- 3})}{\sqrt[]{1 + e^{- 6}}} = \frac{1 + 2 e^{- 3}}{\sqrt[]{1 + e^{- 6}}} = \frac{1 + 2 e^{- 3}}{\sqrt[]{1 + e^{- 6}}} = \frac{e^{3} + 2}{\sqrt[]{e^{6} + 1}}$

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New answer posted

a month ago

Maths Class 11th

If for some m, n; and , then is equal to

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alok kumar singh

Contributor-Level 10

$= 8 * 7 * 6 * 5 * 4 + \frac{9 * 8}{2}$

= 6756

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New answer posted

a month ago

Maths Class 11th

The line passes through the centre of circle x² + y² – 16x – 4y = 0, it interacts with the positive coordinate axis at A & B. Then find the minimum value of OA + OB, where O is origin.

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alok kumar singh

Contributor-Level 10

(y – 2) = m (x – 8)

⇒ x-intercept

⇒ $(\frac{- 2}{m} + 8)$

⇒ y-intercept

⇒ (–8m + 2)

⇒ OA + OB = $\frac{- 2}{m^{2}}$ + 8 – 8m + 2

$f' (m) = \frac{2}{m^{2}} - 8 = 0$

-> $m^{2} = \frac{1}{4}$

-> $m = \frac{- 1}{2}$

-> $f (\frac{- 1}{2}) = 18$

->Minimum = 18

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New answer posted

a month ago

Maths Class 12th

The number of solutions of equation e^{sin x} − 2e^{– sin x} = 2 is

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alok kumar singh

Contributor-Level 10

Take e^sinx = t (t > 0)

⇒ $t - \frac{2}{t} = 2$

⇒ $\frac{t^{2} - 2}{t} = 2$

->t² – 2t – 2 = 0

->t² – 2t + 1 = 3

⇒ (t −1)² = 3

⇒ t = 1 ± $\sqrt[]{3}$

⇒ t = 1 ± 1.73

⇒ t = 2.73 or –0.73 (rejected as t > 0)

⇒ e^{sin x} = 2.73

->log_e e^{sin x} = log_e 2.73

⇒ sin x = log_e 2.73 > 1

So no solution.

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New answer posted

a month ago

Maths Class 11th

Let z₁ and z₂ be two complex numbers such that z₁ + z₂ = 5 and , then the value of is equal to

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alok kumar singh

Contributor-Level 10

z₁ + z₂ = 5

$z_{1}^{3} + z_{2}^{3} = 20 + 15 i$

$z_{1}^{3} + z_{2}^{3} = {(z_{1} + z_{2})}^{3} - 3 z_{1} z_{2} (z_{1} + z_{2})$

$z_{1}^{3} + z_{2}^{3} = 125 - 3 z_{1} \cdot z_{2} (5)$

⇒ 20 + 15i = 125 – 15z₁z₂

⇒ 3z₁z₂ = 25 – 4 – 3i

3z₁z₂ = 21– 3i

z₁⋅z₂ = 7 – i

(z₁ + z₂)² = 25

$z_{1}^{2} + z_{2}^{2} = 25 - 27 (7 - i)$

= 11 + 2i

${(z_{? 1}^{2} + z_{2}^{2})}^{2}$ = 121 − 4 + 44i

⇒ $z_{1}^{4} + z_{2}^{4} + 2 {(7 - i)}^{2} = 117 + 44 i$

⇒ $z_{1}^{4} + z_{2}^{4}$ = 117 + 44i − 2(49 −1−14i )

= 21 + 72i

⇒ $| Z_{1}^{4} + Z_{2}^{4} | = 75$

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New answer posted

a month ago

Maths Class 12th

Let : f(0, R)->∞ be increasing function such that $\underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)} = 1$ then $\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1)$ is equal to

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alok kumar singh

Contributor-Level 10

f is increasing function

x < 5x < 7x

f (x) < f (5x) < f (7x)

$\frac{f (x)}{f (x)} < \frac{f (5 x)}{f (x)} < \frac{f (7 x)}{f (x)}$

$\underset{x \to \infty}{l i m} \frac{f (x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < \underset{x \to \infty}{l i m} \frac{f (7 x)}{f (x)}$

-> $1 < \underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} < 1$ $\underset{x \to \infty}{l i m} \frac{f (5 x)}{f (x)} = 1$

$\underset{x \to \infty}{l i m} (\frac{f (5 x)}{f (x)} - 1) = 0$

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New answer posted

a month ago

Maths Matrices

If $A [\begin{matrix} 1 & 0 & 1 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 1 \end{matrix}]$ , $A [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = 4 [\begin{matrix} - 1 & 0 & 1 \end{matrix}]$ and $A [\begin{matrix} 0 & 1 & 0 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 0 \end{matrix}]$ where, A is a 3 * 3 matrix and (A –3I) $[\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$ then the value of (x, y, z) is

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ .(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 … (6)

–x₃ + z₃ = 4

...more

Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ .(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

⇒ – x₁ + z₁ = −4 … (5)

–x₂ + z₂ = 0 … (6)

–x₃ + z₃ = 4 … (7)

Given $A = [\begin{matrix} 0 & 1 & 0 \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

$[\begin{matrix} y_{1} & y_{2} & y_{3} \end{matrix}] = [\begin{matrix} 0 & 2 & 0 \end{matrix}]$

∴ y₁ = 0, y₂ = 2, y₃ = 0

∴ from (2), (3), (4), (5), (6) and (7)

x₁ = 3, x₂ = 0, x₃ = –1

y₁ = 0, y₂ = 2, y₃ = 0

z₁ = –1, z₂ = 0, z₃ = 3

$A = [\begin{matrix} 3 & 0 & - 1 \\ 0 & 2 & 0 \\ - 1 & 0 & 3 \end{matrix}]$

Now (A – 3I) $[\begin{matrix} x & y & z \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} 0 & 0 & - 1 \\ 0 & - 1 & 0 \\ - 1 & 0 & 0 \end{matrix}] [\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

$[\begin{matrix} - z & - y & - x \end{matrix}]$ = $[\begin{matrix} - 1 & 2 & 3 \end{matrix}]$

[z = 1], [y = –2], [x = –3]

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New answer posted

a month ago

Maths Maths Application of Integrals

The area of the region enclosed by the parabolas y = 4 – x² and 3y = (x – 4)² is in (sq. unit)?

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alok kumar singh

Contributor-Level 10

Area = $| \int_{1}^{4} [{(4 - x)}^{2} - \frac{{(x - 4)}^{2}}{3}] | d x$

Area = $= | (16 - \frac{64}{3}) - (4 - \frac{1}{3} + \frac{27}{9}) |$

$= | (16 - \frac{64}{3}) - (4 - \frac{1}{3} + \frac{27}{9}) |$

$= | 16 - \frac{64}{3} - 4 + \frac{1}{3} + 3 |$

$= | 15 - 2 | = 6$

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