Maths

Let A(a, 0), B(b, 2b + 1) and C(0, b), b $\neq$ 0, $| b | \neq 1,$ be points such that the area of triangle ABC is 1 sq. unit, then the sum of all possible values of a is

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Contributor-Level 10

$a r (Δ A B C) = \frac{1}{2} | \begin{matrix} a & 0 & 1 \\ b & 2 b + 1 & 1 \\ 0 & b & 1 \end{matrix} | = \frac{1}{2} [a (2 b + 1 - b) + (b^{2})] = \frac{1}{2} [a b + a + b^{2}]$

Given ar ( $Δ A B C)$ = 1 so |ab + a + b²| = 2

$\Rightarrow a (b + 1) + b^{2} = \pm 2$

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

So sum = $\frac{- 2 b^{2}}{b + 1}$

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New answer posted

a year ago

The distance of the point (1, -2, 3) form the plane x – y + z = 5 measured parallel to a line, whose direction ratios are 2, 3, -6 is

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Contributor-Level 10

Equation of line PQ :

$\frac{x - 1}{2} = \frac{y + 2}{3} = \frac{z - 3}{- 6} = k$

So, let Q (2k + 1, 3k – 2, -6k + 3) Q lies on given plane so

2k + 1 – 3k + 2 – 6k + 3 = 5

$- 7 k = - 1 o r k = \frac{1}{7}$

So, $Q (\frac{9}{7}, \frac{- 11}{7}, \frac{15}{7})$

Now required distance = PQ = $\sqrt[]{\frac{4}{49} + \frac{9}{49} + \frac{36}{49}} = 1$

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New answer posted

a year ago

A tangent and a normal are drawn at the point P(2, -4) on the parabola y² = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is eqal to :

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Contributor-Level 10

Equation of tangent at P (2, -4)

y (-4) = 4 (x + 2)

x + y + 2 = 0

So, A (-2, 0)

Equation of normal at P:

y + 4 = 1 (x – 2)

x – y = 6

So, B (-2, -8)

For square mid-point of AB = mid-point of PQ

$\Rightarrow \frac{a + 2}{2} = - 2 \Rightarrow a = - 6$

$\frac{b - 4}{2} = \frac{- 8}{2} \Rightarrow b = - 4$

So, 2a + b = -16

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New answer posted

a year ago

Let $\frac{s i n A}{s i n B} = \frac{s i n (A - C)}{s i n (C - B)},$ where A, B, C are angles of a triangle ABC. If the lengths of the sides opposite these angles are a, b, c respectively, then:

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Contributor-Level 10

$\frac{s i n A}{s i n B} = \frac{s i n (A - C)}{s i n (C - B)}$

sin A sin C cos B – sin A cos C sin B = sin B sin A cos C – sin B cos A sin C

2 sin A sin B cos C = sin A sin C cos B + sin B sin C cos A

By sine rule $\frac{s i n A}{a} = \frac{s i n B}{b} = \frac{s i n C}{c} = k$

$2 . a k . b k . \frac{(a^{2} + b^{2} - c^{2})}{2 a b} = a k . c k . \frac{(c^{2} + a^{2} - b^{2})}{2 a c} + b k . c k . \frac{(b^{2} + c^{2} - a^{2})}{2 b c}$

$\Rightarrow b^{2}, c^{2}, a^{2} a r e i n A . P .$

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New answer posted

a year ago

If two tangent from a point P to the parabola y² = 16(x – 3) are at right angles, then the locus of point P is:

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Contributor-Level 10

Locus of point of intersection of perpendicular tangent will be its director circle & director circle of parabola be its directrix.

Given parabola y² = 16 (x – 3) so equation of directrix be x – 3 = - 4 i.e., x + 1 = 0

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New answer posted

a year ago

A wire of length 20m is to be cut into two pieces. One of the pieces is to be made into a square and the other into a regular hexagon. Then the length of the side (in meters) of the hexagon, so that the combined area of the square and the hexagon is minimum, is:

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Contributor-Level 10

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a = \frac{x}{4} & b = \frac{y}{6}$

Now let A = area of square + area of hexagon

$A = \frac{x^{2}}{16} + 6 * \frac{\sqrt[]{3}}{4} * \frac{y^{2}}{36} = \frac{x^{2}}{16} + \frac{\sqrt[]{3} y^{2}}{24} = \frac{x^{2}}{16} + \frac{\sqrt[]{3}}{24} {(20 - x)}^{2} f r o m (i)$ ,

for minimum area

$x = \frac{80 \sqrt[]{3}}{6 + 4 \sqrt[]{3}} = \frac{80 \sqrt[]{3}}{2 \sqrt[]{3} (\sqrt[]{3} + 2)} = \frac{40}{2 + \sqrt[]{3}}$

$x = 40 (2 - \sqrt[]{3})$

=> side of hexagon = $\frac{y}{6} = \frac{20 \sqrt[]{3} (2 - \sqrt[]{3})}{6} = \frac{20 \sqrt[]{3}}{6 (2 + \sqrt[]{3})} = \frac{10 \sqrt[]{3}}{3 (2 + \sqrt[]{3})} = \frac{10}{2 \sqrt[]{3} + 3}$

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New answer posted

a year ago

If y(x) = cot^-1 $(\frac{\sqrt[]{1 + s i n x} + \sqrt[]{1 - s i n x}}{\sqrt[]{1 + s i n x} - \sqrt[]{1 - s i n x}}), x \in (\frac{π}{2}, π), t h e n \frac{d y}{d x} a t x = \frac{5 π}{6} i s :$

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Contributor-Level 10

$y (x) = c o t^{- 1} (\frac{\sqrt[]{1 + s i n x} + \sqrt[]{1 - s i n x}}{\sqrt[]{1 + s i n x} - \sqrt[]{1 - s i n x}}), x \in (\frac{π}{2}, π)$

$= c o t^{- 1} (\frac{s i n \frac{x}{2} + c o s \frac{x}{2} + s i n \frac{x}{2} - c o s \frac{x}{2}}{s i n \frac{x}{2} + c o s \frac{x}{2} - s i n \frac{x}{2} + c o s \frac{x}{2}}) = c o t^{- 1} t a n \frac{x}{2} = c o t^{- 1} c o t (\frac{π}{2} - \frac{π}{2}) = \frac{π}{2} - \frac{x}{2}$

$\frac{d y}{d x} = \frac{- 1}{2}$

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New answer posted

a year ago

Let A = $([\begin{matrix} [x + 1] & [x + 2] & [x + 2] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix}])$ where [t] denotes the greatest integer less than or equal to t. If det(A) = 192, then the set of values of x is the interval.

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Contributor-Level 10

$| A | = | \begin{matrix} [x + 1] & [x + 2] & [x + 3] \\ [x] & [x + 3] & [x + 3] \\ [x] & [x + 2] & [x + 4] \end{matrix} | = | \begin{matrix} [x] + 1 & [x] + 2 & [x] + 3 \\ [x] & [x] + 3 & [x] + 3 \\ [x] & [x] + 2 & [x] + 4 \end{matrix} |$

$R_{1} \to R_{1} - R_{3} & R_{2} \to R_{2} - R_{3}$

$| \begin{matrix} 1 & 0 & - 1 \\ 0 & 1 & - 1 \\ [x] & [x] + 2 & [x] + 4 \end{matrix} | = 192$

$\Rightarrow [x] = 62$

$\Rightarrow x \in [62, 63)$

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New answer posted

a year ago

If $y^{1 / 4} + y^{- 1 / 4} = 2 x a n d (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + α x \frac{d y}{d x} + β y = 0, t h e n | α - β |$ is equal to_________

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Contributor-Level 10

$y^{1 / 4} + \frac{1}{y^{1 / 4}} = 2 x$

${(y^{1 / 4})}^{2} - 2 x y^{1 / 4} + 1 = 0$

=> ${(y^{1 / 4})}^{2} - 2 x y^{1 / 4} + 1 = 0$

$\Rightarrow \frac{d y}{d x} = \frac{4 y}{\sqrt[]{x^{2} - 1}} . . . . . . . . . . (i)$

$\Rightarrow (x^{2} - 1) \frac{d^{2} y}{d x^{2}} + \frac{x d y}{d x} - 16 y = 0 f r o m (i)$

=>α = 1, β = -16

|α - β| = 17

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New answer posted

a year ago

The set of all values of k > -1, for which the equation ${(3 x^{2} + 4 x + 3)}^{2} - (k + 1) (3 x^{2} + 4 x + 3) (3 x^{2} + 4 x + 2) + k {(3 x^{2} + 4 x + 2)}^{2} = 0$ has real roots is:

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