Conic Sections

If the tangents drawn at the points P and Q on the parabola y² = 2x – 3 intersect at the point R(0, 1), then the orthocenter of the triangle PQR is:

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Contributor-Level 10

$y^{2} = 2 x - 3$ ……. (i)

Equation of chord of contact

PQ : r = O

(y * 1) = (x + 0) – 3

y = x – 3 ……… (ii)

From (i) and (ii)

y = 1 or 3

$M P Q = \frac{4}{4} = 1$

$M Q R = \frac{2}{6} = \frac{1}{3}$

$M_{P Q} * M_{P R} = - 1 \Rightarrow P Q ⊥ P R$

Orthocentre = P (2, -1)

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2 months ago

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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alok kumar singh

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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2 months ago

If the length of the latus rectum of the ellipse x² + 4y² + 2x + 3y $- λ = 0$ is 4, and $l$ is the length of the its major axis, then $λ + l$ is equal to

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$\frac{{(x + 1)}^{2}}{λ + 5} = \frac{{(y + 1)}^{2}}{\frac{λ + 5}{4}} = 1$

length of latus rectum = $\frac{2 b^{2}}{a} = \frac{2 (\frac{λ + 5}{4})}{\sqrt[]{5 + λ}} = 4$

$\sqrt[]{λ + 5} = 8 \Rightarrow λ = 59$

Major axis = $2 \sqrt[]{λ + 5} = 16$

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2 months ago

An ellipse $E : \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through the vertices of the hyperbola $H : \frac{x^{2}}{49} - \frac{y^{2}}{64} = - 1 .$ Let the major and minor axes of the ellipse E coincide with transverse and conjugate axes of the hyperbola H, respectively. Let the product of the eccentricities of E and H be $\frac{1}{2} .$ If $l$ is the length of the latus rectum of the ellipse E, then the value of 113 $l$ is equal to

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Contributor-Level 10

$e_{H} = \sqrt[]{1 + \frac{64}{49}} = \frac{\sqrt[]{113}}{7}$

$\Rightarrow e_{H} . e_{E} = \frac{1}{2}$

$\Rightarrow \frac{113}{49} . \frac{(64 - a^{2})}{64} = \frac{1}{4} \Rightarrow a^{2} - 64 = \frac{3 2^{2}}{113}$

$l = \frac{2 a^{2}}{b} = 2 (64 + \frac{3 2^{2}}{113}) . \frac{1}{8}$

$113 l = 1552$

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2 months ago

If the circles x² + y² + $2 (3 - \sqrt[]{3})$ x + $2 (4 - \sqrt[]{6}) y = k + 6 \sqrt[]{3} + 8 \sqrt[]{6}, k > 0,$ touch internally at the point P(, ), then ${(α + \sqrt[]{3})}^{2} + {(β + \sqrt[]{6})}^{2}$ is equal to

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Contributor-Level 10

The circle $x^{2} + y^{2} + 6 x + 8 y + 16 = 0$ has centre (3, 4) and radius 3 units

The circle

$x^{2} + y^{2} + 2 (3 - \sqrt[]{3}) x + 2 (4 - \sqrt[]{6}) y = k + 6 \sqrt[]{3} + 8 \sqrt[]{6}, k > 0$ has centre $(\sqrt[]{3} - 3, \sqrt[]{6} - 4)$ and radius $\sqrt[]{k + 34}$

$?$ These two circles touch internally hence

$\sqrt[]{3 + 6} = | \sqrt[]{k + 34} - 3 |$

here, k = 2 is only possible $(? k > 0)$

Equation of common tangent to two circle is

$2 \sqrt[]{3} x + 2 \sqrt[]{6} y + 16 + 6 \sqrt[]{3} + 8 \sqrt[]{6} + k = 0$

$? k = 2$ then equation is

$x + \sqrt[]{2} y + 3 + 4 \sqrt[]{2} + 3 \sqrt[]{3} = 0$ ….(i)

$? (α, β)$ are foot of perpendicular from (3, 4) to line (i) then

$\frac{α + 3}{1} = \frac{β + 4}{\sqrt[]{2}} = \frac{- 3 - 4 \sqrt[]{2} + 3 + 4 \sqrt[]{2} + 3 + \sqrt[]{3}}{1 + 2}$

$∴ {(α + \sqrt[]{3})}^{2} + {(β + \sqrt[]{6})}^{2} = 25$

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2 months ago

The sum of diameters of the circles that touch (i) the parabola 75x² = 64(5y – 3) at the point $(\frac{8}{5}, \frac{6}{5})$ and (ii), the y-axis, is equal to………….

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Contributor-Level 10

Equation of tangent to the parabola at

$P (\frac{8}{5}, \frac{6}{5})$

$75 x . \frac{8}{5} = 160 (y + \frac{6}{5}) - 192$

$\Rightarrow 120 x = 160 y$

$\Rightarrow 3 x = 4 y$

Equation of circle touching the given parabola at P can be taken as

$\Rightarrow λ = \frac{2}{5} o r - \frac{8}{5}$

Radius = 1 or 4

Sum of diameter = 10

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2 months ago

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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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

2 months ago

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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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

2 months ago

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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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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2 months ago

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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