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Conic Sections

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6 months ago

Conic Sections Class 11th

Le the equation of two diameters of a circle x² + y² -2x + 2fy + 1 = 0 be 2px – y = 1 and 2x + py = 4p. Then the slope $m \in (0, \infty)$ of the tangent to the hyperbola 3x² – y² = 3 passing through the centre of the circle is equal to………….

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Vishal Baghel

Contributor-Level 10

$x^{2} + y^{2} - 2 x + 2 f y + 1 = 0 c e n t r e = (1, - f)$

diameter 2px – y = 1 ………. (i)

2x + py = 4p ……… (ii)

$x = \frac{5 P}{2 P^{2} + 2}$

f = 0

$[f o r P = \frac{1}{2}]$

$\frac{5 P}{2 P^{2} + 2} = 1$

f = 3 [for P = 2]

substitute (2, 3)

$3 = m \pm \sqrt[]{m^{2} - 3}$

$∴ m = 2$

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New answer posted

6 months ago

Conic Sections Class 11th

Let AB be a chord of length 12 of the circle ${(x - 2)}^{2} + {(y + 1)}^{2} = \frac{169}{4} .$

If tangents drawn to the circle at points A and B intersect at the point P, then five times the distance of point P form chord AB is equal to……………

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Payal Gupta

Contributor-Level 10

Let C be the centre and M be the mid point of AB

$Δ A P C : s i n θ = \frac{13 / 2}{p c} = \frac{5}{13} \Rightarrow p C = \frac{169}{10}$

$Δ A M C : c o s θ = \frac{6}{13 / 2} = \frac{12}{13}$

PC = $\frac{169}{10}, M C = \frac{13}{2} s i n θ = \frac{13}{2} \cdot \frac{5}{13}$

PM = PC – MC = $\frac{169}{10} - \frac{5}{2} = \frac{144}{10}$

5PM = 72

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New answer posted

6 months ago

Conic Sections Class 11th

For the hyperbola H: x² – y² = 1 and the ellipse E : $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ a > b > 0, let the

(1) eccentricity of E be reciprocal of the eccentricity of H, and

(2) the line y = $\sqrt[]{\frac{5}{2}} x + K$ be a common tangent of E and H.

Then 4(a² + b²) is equal to…………….

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Vishal Baghel

Contributor-Level 10

$e_{E} = \sqrt[]{1 - \frac{b^{2}}{a^{2}}}, e_{H} = \sqrt[]{2}$

$I f \Rightarrow e_{E} = \frac{1}{e_{H}} \Rightarrow \frac{a^{2} - b^{2}}{a^{2}} = \frac{1}{2}$

$k^{2} = a^{2} * \frac{5}{2} + b^{2} = \frac{3}{2}$

$6 b^{2} = \frac{3}{2} \Rightarrow b^{2} = \frac{1}{4} a n d a^{2} = \frac{1}{2}$

$∴ 4 (a^{2} + b^{2}) = 3$

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New answer posted

6 months ago

Conic Sections Class 11th

Let the abscissae of the two points P and Q be the roots of 2x² – rx + p = 0 and the ordinates of P and Q be the roots of x² – sx – q = 0. If the equation of the circle described on PQ as diameter is 2(x² + y²) – 11x – 14y – 22 = 0, then 2r + s – 2q + is equal to….

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Vishal Baghel

Contributor-Level 10

$P (x_{1}, y_{1})$

$Q (x_{2}, y_{2})$

$x_{1} + x_{2} = \frac{r}{2}, x_{1} x_{2} = \frac{P}{2}$

$y_{1} + y_{2} = s, y_{1}, y_{2} = - 9$

$(x - x_{1}) (x - x_{2}) + (y - y_{1}) (y - y_{2}) = 0$

$2 (x^{2} + y^{2}) - r x - 2 s y + p - 2 q = 0$

r = 11, s = 7, p – 2q = -22

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New answer posted

6 months ago

Conic Sections Class 11th

If the solution curve y = y(x) of the differential equation y²dx + (x² – xy + y²)dy = 0, which passes through the point (1, 1) and intersects the line y = $\sqrt[]{3} x$ at the point $(α, \sqrt[]{3} α),$ then value of $l o g_{e} (\sqrt[]{3} α)$ is equal to:

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Vishal Baghel

Contributor-Level 10

$y^{2} d x + (x^{2} ? x y + y^{2}) d y = 0$

$\frac{d x}{d y} + \frac{x^{2} ? x y + y^{2}}{y^{2}} = 0 ? \frac{d x}{d y} + {(\frac{x}{y})}^{2} ? (\frac{x}{y}) + 1 = 0$

Put x = vy $? v + y \frac{d v}{d y} + v^{2} ? v + 1 = 0$

$? \frac{y d v}{d y} + v^{2} + 1 = 0$

$? \frac{?}{6} + l n | y | = \frac{?}{4}$

$l n | y | = \frac{?}{12}$

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New answer posted

6 months ago

Conic Sections Class 11th

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Vishal Baghel

Contributor-Level 10

$y^{2} d x + (x^{2} ? x y + y^{2}) d y = 0$

$\frac{d x}{d y} + \frac{x^{2} ? x y + y^{2}}{y^{2}} = 0 ? \frac{d x}{d y} + {(\frac{x}{y})}^{2} ? (\frac{x}{y}) + 1 = 0$

Put x = vy $? v + y \frac{d v}{d y} + v^{2} ? v + 1 = 0$ $\frac{}{}^{}$

$? \frac{y d v}{d y} + v^{2} + 1 = 0$

$? \frac{?}{6} + l n | y | = \frac{?}{4}$

$l n | y | = \frac{?}{12}$

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New answer posted

6 months ago

Conic Sections Class 11th

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Vishal Baghel

Contributor-Level 10

$y^{2} d x + (x^{2} ? x y + y^{2}) d y = 0$

$\frac{d x}{d y} + \frac{x^{2} ? x y + y^{2}}{y^{2}} = 0 ? \frac{d x}{d y} + {(\frac{x}{y})}^{2} ? (\frac{x}{y}) + 1 = 0$

Put x = vy $? v + y \frac{d v}{d y} + v^{2} ? v + 1 = 0$ $\frac{}{}^{}$

$? \frac{y d v}{d y} + v^{2} + 1 = 0$

$? \frac{?}{6} + l n | y | = \frac{?}{4}$

$l n | y | = \frac{?}{12}$

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New answer posted

6 months ago

Conic Sections Class 11th

Let x = 2t, $y = \frac{t^{2}}{3}$ be a conic. Let S be the focus and B the point on the axis of the conic such that $S A ⊥ B A,$ where A is any point on the conic. If k is the ordinate of the centroid of the $Δ S A B,$ then $\underset{t \to 1}{l i m}$ k is equal to:

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

x = 2t $A (2 t, \frac{t^{2}}{3})$ $\frac{^{}}{}$

$\frac{^{}}{}$ $y = \frac{t^{2}}{3}$ S (0, 3)

$3 y = {(\frac{x}{2})}^{2}$ $B (0, ?)$

$3 k = \frac{t^{2}}{3} + 3 + ? = \frac{2 t^{2}}{3} + 3 ? \frac{12 t^{2}}{9 ? t^{2}}$

$\underset{t ? 1}{l i m} 3 k = \frac{2}{3} + 3 ? \frac{12}{8} = 3 ? \frac{5}{6} = \frac{13}{6}$

${\frac{}{}}^{}$

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New answer posted

6 months ago

Conic Sections Class 11th

If y = m₁x + c₁ and y = m₂x + c₂, m₁ $\neq$ m₂ are two common tangents of circle x² + y² = 2 and parabola y² = x, Then the value of $8 | m_{1} m_{2} |$ is equal to:

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Vishal Baghel

Contributor-Level 10

$y^{2} = 4 (\frac{1}{4}) x$

^{$x - t y + \frac{1}{4} t^{2} = 0$}

^{$\frac{1}{t} = m, m_{2} \Rightarrow (8 m_{1} m_{2}) = 3 \sqrt[]{2} - 4$}

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New answer posted

6 months ago

Conic Sections Class 11th

If two tangents drawn from a point (a, b) lying on the ellipse 25x² + 4y² = 1 to the parabola y² = 4x are such that the slope of one tangent is four times the other, then the value of ${(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2}$ equals __________.

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alok kumar singh

Contributor-Level 10

$\frac{x^{2}}{\frac{1}{25}} + \frac{y^{2}}{\frac{1}{4}} = 1$

$25 α^{2} + 4 β^{2} = 1 . . . . . . . . . . (i)$

Equation of tangent to parabola y = mx + $\frac{1}{m}$ passes

though (a, b)

$α m^{2} - β m + 1 = 0$

$m_{1} + m_{2} = \frac{β}{α}, 4 m_{1}^{2} = \frac{1}{α}$

$∴ f r o m (i) & (i i)$

25(a² + a) = 1 …………(iii)

$∴ {(10 α + 5)}^{2} + {(16 β^{2} + 50)}^{2} = 2929$

$∴ a = 1 + 0 + b {(s i n - \frac{π}{2})}^{\frac{π}{2}} + 2 b + 1$

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