Conic Sections

If the minimum area of the triangle formed by a tangent to the ellipse $\frac{x^{2}}{b^{2}} + \frac{y^{2}}{4 a^{2}} = 1$ and the co-ordinate axis is kab, then k is equal to________

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Equation of tangent to given ellipse at

$P : \frac{x . c o s θ}{b} + \frac{y . s i n θ}{2 a} = 1$

A (b sec θ. 0) 7 B (0, 2a cosec θ)

area of $Δ O A B$

$= \frac{2 a b}{2 s i n θ c o s θ} = \frac{2 a b}{s i n 2 θ}$

For minimum area sin 2θ = 1

So minimum area = 2ab

=>k = 2

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6 months ago

Let the equation $x^{2} + y^{2} + p x + (1 - p) y + 5 = 0$ represent circles of varying radius r $\in$ (0, 5]. Then the number of elements in the set S = {q : q = p² and q is an integer) is________

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$r = \sqrt[]{{(\frac{p}{2})}^{2} + {(\frac{1 - p}{2})}^{2} - 5} = \sqrt[]{\frac{p^{2} + 1 + p^{2} - 2 p - 20}{4}} = \sqrt[]{\frac{2 p^{2} - 2 p - 19}{2}}$

Since $r \in (0, 5) s o 0 < \sqrt[]{2 p^{2} - 2 p - 19} < 10$

$\Rightarrow 2 p^{2} - 2 p - 19 > 0 & 2 p^{2} - 2 p - 19 < 100$

$P \in [\frac{1 - \sqrt[]{239}}{2}, \frac{1 - \sqrt[]{39}}{2}) \cup (\frac{1 + \sqrt[]{39}}{2}, \frac{1 + \sqrt[]{239}}{2}]$

So number of integral values of P² is 61.

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6 months ago

The point $P (- 2 \sqrt[]{6}, \sqrt[]{3})$ lies on the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ having eccentricity $\frac{\sqrt[]{5}}{2} .$ If the tangent and normal at P to the hyperbola intersect its conjugate axis at the points Q and R respectively, then QR is equal to:

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$P (- 2 \sqrt[]{6}, \sqrt[]{3}) l i e s o n \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{24}{a^{2}} - \frac{3}{b^{2}} = 1 . . . . . . . . . (i)$

$b^{2} = \frac{a^{2}}{4} . . . . . . . . . (i i)$

solving (i) & (ii) a² = 12 Þ b² = 3 hyperbola $\frac{x^{2}}{12} - \frac{y^{2}}{3} = 1$

Cuts conjugate axis at $R (0, R \sqrt[]{3})$

$∴ Q R = 6 \sqrt[]{3}$

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6 months ago

A circle C touches the line x = 2y at the point (2, 1) and intersects the circle C₁ : x² + y² + 2y – 5 = 0 at two point P and Q such that PQ is a diameter of C₁. Then the diameter of C is

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Equation of required circle

$C : {(x - 2)}^{2} + {(y - 1)}^{2} + λ (2 y - x) = 0 . . . . . . . . . . (i)$

Intersect C₁x² + y² + 2y – 5 = 0 .(ii)

$∴$ Equation of radical axis

$- 4 x - 4 y + 10 + λ (2 y - x) = 0 . . . . . . . . . . (i i)$

Centre of C₁ (0, -1) lies on .(ii)

$\Rightarrow 4 + 10 - 2 λ = 0 \Rightarrow λ = 7$

Equation of circle C is

$x^{2} + y^{2} - 11 x + 12 y + 5 = 0$

Diameter = $\sqrt[]{245} = 7 \sqrt[]{5}$

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6 months ago

If a ling along a chord of the circle 4x² + 4y² + 120x + 675 = 0, passes through of the point (-30, 0) and is tangent to the parabola y² = 30x, then the length of this chord is:

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$P (\frac{15}{2} t^{2}, 15 t)$

C (-15,0)

Q (-30,0)

tanθ = $\frac{P N}{Q N}$

=> $\frac{1}{t} = \frac{15 t}{30 + \frac{15}{2} t^{2}}$

=> t = 2

$T a n g e n t = y = \frac{1}{2} (x + 30)$

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6 months ago

On the ellipse $\frac{x^{2}}{8} + \frac{y^{2}}{4} = 1$ let P be a point in the second quadrant such that the tangent at P to the ellipse is perpendicular to the line x + 2y = 0. Let S and S' be the foci of the ellipse and e be its eccentricity. If A is the area of the triangle SPS' then, the value of (5 – e²). A is:

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$T_{(P)} : \frac{x x_{1}}{8} + \frac{y y_{1}}{4} = 1,$

where $\frac{x_{1} 2}{8} + \frac{y_{1} 2}{4} = 1 . . . . . . . . . . . (i)$

given : slope = 2

=> x₁ = -4y . (ii)

(i) & (ii) => P $(- \frac{8}{3}, \frac{2}{3})$

$e = \frac{1}{\sqrt[]{2}}$

A = ar (PSS') = $\frac{4}{3}$

$(5 - e^{2}) A = 6$

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6 months ago

The locus of a point which moves such that the sum of squares of its distances from the points (0, 0), (1, 0), (0, 1), (1, 1) is 18 units, is a circle of diameter d. Then d² is equal to_______.

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$x^{2} + y^{2} + {(x - 1)}^{2} + y^{2} + x^{2} + {(y - 1)}^{2} + {(x - 1)}^{2} + {(y - 1)}^{2} = 18$

$\Rightarrow x^{2} + y^{2} - x - y - \frac{7}{2} = 0$

$r^{2} = \frac{1}{4} + \frac{1}{4} + \frac{7}{2} = 4$

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6 months ago

The locus of the mid points of the chords of the hyperbola x² – y² = 4, which touch the parabola y² = 8x, is:

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Let P (h, k) be the mid point of the chord x² – y² = 4

$∴$ its equation is xh – yk = h² – k²

$O r, y = (\frac{h}{x}) x + \frac{k^{2} - h^{2}}{k}$ if this line is tangent to y² = 8x then $\frac{k^{2} - h^{2}}{k}$ = $\frac{2}{h / k} = \frac{2 k}{h}$

$h (k^{2} - h^{2}) = 2 k^{2}$

$∴$ Required locus is 2y² = x (y² – x²)

$\Rightarrow x^{3} = y^{2} (x - 2)$

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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:

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

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

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

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

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

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6 months ago

Let the eccentricity of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 b e \frac{5}{4} .$ If the equation of the normal at the point $(\frac{8}{\sqrt[]{5}}, \frac{12}{5})$ on the hyperbola is $8 \sqrt[]{5} x + β y = λ, t h e n λ - β$ is equal to…………….

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