Ncert Solutions Maths class 11th NCERT Maths Spl Conic Sections

What are the key formulas in Conic Sections?

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Those who are preparing for any competive exam such as JEE Main, NDA and others really need to memorise formulas. Check the important formulas below;

Circle

Standard equation (center at origin): $x^{2} + y^{2} = r^{2}$
$x^2 + y^2 = r^2$
General form: $x^2 + y^2 + 2gx + 2fy + c = 0$
Center = (–g, –f), Radius = ?(g² + f² – c)

Parabola:

Standard form (horizontal axis): $y^{2} = 4 a x$
$y^2 = 4ax$
Standard form (vertical axis): $x^{2} = 4 a y$
$x^2 = 4ay$
Vertex: (0, 0)
Focus: (a, 0) or (0, a)
Directrix: x = –a or y = –a
Latus rectum = 4a

Ellipse: Check all the important topics related to this topic below;

Standard form (major axis along x-axis):
- $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where a > b
- Eccentricity: $e = \sqrt{1 ? \frac{b^{2}}{a^{2}}}$
  $e = \sqrt{1 - \frac{b^2}{a^2}}$

Foci: (±ae, 0)
Latus rectum: $\frac{2b^2}{a}$

Hyperbola: Students can check impor

...more

Those who are preparing for any competive exam such as JEE Main, NDA and others really need to memorise formulas. Check the important formulas below;

Circle

Standard equation (center at origin): $x^{2} + y^{2} = r^{2}$
$x^2 + y^2 = r^2$
General form: $x^2 + y^2 + 2gx + 2fy + c = 0$
Center = (–g, –f), Radius = ?(g² + f² – c)

Parabola:

Standard form (horizontal axis): $y^{2} = 4 a x$
$y^2 = 4ax$
Standard form (vertical axis): $x^{2} = 4 a y$
$x^2 = 4ay$
Vertex: (0, 0)
Focus: (a, 0) or (0, a)
Directrix: x = –a or y = –a
Latus rectum = 4a

Ellipse: Check all the important topics related to this topic below;

Standard form (major axis along x-axis):
- $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where a > b
- Eccentricity: $e = \sqrt{1 ? \frac{b^{2}}{a^{2}}}$
  $e = \sqrt{1 - \frac{b^2}{a^2}}$

Foci: (±ae, 0)
Latus rectum: $\frac{2b^2}{a}$

Hyperbola: Students can check important formulae related to this topic below;

Standard form (transverse axis along x-axis):
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
Eccentricity:
$e = \sqrt{1 + \frac{b^2}{a^2}}$
Foci: (±ae, 0)
Latus rectum: $\frac{2b^2}{a}$
Asymptotes:
$y = \pm \frac{b}{a} x$

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Similar Questions for you

Let $H : \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, a > 0, b > 0,$ be a hyperbola such that the sum of lengths of the transverse and the conjugate axes is $4 (2 \sqrt[]{2} + \sqrt[]{14}) .$ If the eccentricity H is $\frac{\sqrt[]{11}}{2},$ then the value of a² + b² is equal to………….

alok kumar singh

Given 2a + 2b = $4 (2 \sqrt[]{2} + \sqrt[]{14})$

$= 4 \sqrt[]{2} (2 + \sqrt[]{7})$

$b^{2} = a^{2} (\frac{11}{4} - 1)$

4b2 = 7a2

b = $\frac{\sqrt[]{7}}{2} a$

Let the tangent to the circle C₁ : x² + y² = 2 at the point M(1, 1) intersect the circle $C_{2} : {(x - 3)}^{2}$ + ${(y - 2)}^{2} = 5,$ at two distinct points A and B. If the tangents to C₂at the points A and B intersect at N, then the area of the triangle ANB is equal to:

alok kumar singh

Tangent to C₁ at (-1, 1) is T = 0

x(-1) + 4(1) = 2

-x + y = 2

find OP by dropping from (3, 2) to centre

OP = $| \frac{3 - 2 + 2}{\sqrt[]{2}} | = \frac{3}{\sqrt[]{2}}$

AP = $\sqrt[]{r^{2} - O P^{2}}$

$= \sqrt[]{5 - \frac{9}{2}} = \frac{1}{\sqrt[]{2}}$

$t a n θ = \frac{O P}{A P} = \frac{3 / \sqrt[]{2}}{1 / \sqrt[]{2}} = 3$

area of $Δ A B N = \frac{1}{2} A N^{2} \cdot s i n 2 θ$

AN = $\frac{\sqrt[]{5}}{3}$

$= \frac{1}{2} \cdot \frac{5}{9} \cdot (\frac{2 t a n θ}{1 + t a n^{2} θ})$

$= \frac{5}{2.9} \times \frac{2.3}{1 + 3^{2}} = \frac{1}{6}$

sin = $\frac{A P}{A N}$

Let PQ be a focal chord of the parabola y2 = 4x such that it subtends an angle of $\frac{π}{2}$ at the point (3, 0). Let the line segment PQ be also a focal chord of the ellipse E: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1,$ $a^{2} > b^{2} .$ If e is the eccentricity of the ellipse E, then the value of $\frac{1}{e^{2}}$ is equal to:

alok kumar singh

$A P ⊥ B P$

M₁ M₂ = 1

$\frac{2 t}{t^{2} - 3} \times \frac{2 / 6}{3 - \frac{1}{t^{2}}} = - 1$

t = 1

So, A (1, 2) and B (1, 2) they must be end pts of focal chord.

Length of latus rectum $= \frac{2 b^{2}}{a}$

$4 = \frac{2 b^{2}}{a}$

b2 = 2a and ae = 1

Eccentricity of ellipse (Horizontal)

b²= a² (1 – e²)

2a = a² (1 – e²)

2 = $\frac{1}{e} (1 - e^{2})$

e²+ 2e – 1 = 0

$e = \frac{- 2 \pm \sqrt[]{4 + 4}}{2}$

$e = - 1 + \sqrt[]{2}$

now $\frac{1}{e^{2}}$ $=$ $3$ $+$ $2$

If the line x – 1 = 0 is a directrix of the hyperbola kx² – y² = 6, then the hyperbola passes through the point

alok kumar singh

Given hyperbola : $\frac{k x^{2}}{6} - \frac{y^{2}}{6} = 1$ so eccentricity e = $\sqrt[]{1 + k}$ and directrices $x = \pm \frac{a}{e}$

$\Rightarrow x = \pm \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} \Rightarrow \frac{\sqrt[]{6}}{\sqrt[]{k} \sqrt[]{k + 1}} = 1$

k = 2 therefore equation of hyperbola is $\frac{x^{2}}{3} - \frac{y^{2}}{6} = 1$

hence it passes through the point $(\sqrt[]{5}, - 2)$

Let the abscissae of two points P and Q on a circle be the roots of x² – 4x – 6 = 0 and the ordinates of P and Q be the roots of y² + 2y – 7 = 0. If PQ is a diameter of the circle x² + y² + 2ax + 2by + c = 0, then the value of (a + b +) is……………

alok kumar singh

Abscissae of PQ are roots of x² – 4x – 6 = 0

Ordinates of PQ are roots of y² + 2y – 7 = 0 and PQ is diameter

$∴$ Equation of circle is $x^{2} + y^{2} - 4 x + 2 y - 13 = 0$ ……………. (i)

But, given $x^{2} + y^{2} + 2 a x + 2 b y + c = 0$ ……………. (ii)

By comparison a = -2, b = 1, c = -13 Þ a + b – c = -2 + 1 + 13 = 12

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