Tags Conic Sections

Conic Sections

Get insights from 199 questions on Conic Sections, answered by students, alumni, and experts. You may also ask and answer any question you like about Conic Sections

Follow Ask Question

199

Questions

Discussions

Active Users

Followers

New answer posted

2 months ago

Conic Sections Class 11th

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………….

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

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 = $2 \sqrt[]{λ + 5} = 16$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

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……….

0 Follower 2 Views

alok kumar singh

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

0 0

New answer posted

2 months ago

Conic Sections Class 11th

If the circle x² + y² – 2gx + 6y – 19c = 0, g, c $\in$ R passes through the point (6, 1) and its centre lies on the line x – 2cy = 8, then the length of intercept made by the circle on x-axis is

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

Circle passes through (6, 1)

12 g – 19 c = 43 …. (i)

Centre lies on x – 2xy = 8

->g + 6c = 8 …. (ii)

From (i) & (ii), c = 1, 9 = 2

Length of x – intercept - $2 \sqrt[]{g^{2} - C}$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

Let P(a, b) be a point on the parabola y² = 8x such that the tangent at P passes through the centre of the circle x² + y² – 10x – 14y + 65 = 0. Let A be the product of all possible values of a and B be the product of all possible values of b. Then the value of A + B is equal to

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

tangent at (2t², 4t) is ty = x + 2t²,

It passes through (5, 7)

$2 t^{2} - 7 t + 5 = 0 \Rightarrow t = 1, \frac{5}{2}$

$P (2 t^{2}, 4 t)$ will be (2, 4), $(\frac{25}{2}, 0)$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola y = 2x and touches the parabola $y = {(x - \frac{1}{4})}^{2} + α,$ where a > 0. Then (4a - 8)² is equal to………….

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Let the equation of circle be

$x (x - \frac{1}{2}) + y^{2} + λ y = 0$

$\Rightarrow x^{2} + y^{2} - \frac{1}{2} x + λ y = 0$

Radius = $\sqrt[]{\frac{1}{16} + \frac{λ^{2}}{4}} = 2$

$\Rightarrow λ^{2} = \frac{63}{4} \Rightarrow {(x - \frac{1}{4})}^{2} + {(y + \frac{λ}{2})}^{2} = 4$

$?$ This circle and parabola

$y - α = {(x - \frac{1}{4})}^{2}$ touch each other, so

$α = - \frac{λ}{2} + 2 \Rightarrow α - 2 = - \frac{λ}{2} \Rightarrow {(α - 2)}^{2} = \frac{λ^{2}}{4} = \frac{63}{16}$

${(4 α - 8)}^{2} = 63$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

Let the eccentricity of an ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, a > b, b e \frac{1}{4} .$ If this ellipse passes through the point $(- 4 \sqrt[]{\frac{2}{5}}, 3)$ , then a² + b² is equal to:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$

$\Rightarrow \frac{{(- 4 \sqrt[]{\frac{2}{5}})}^{2}}{a^{2}} + \frac{32}{b^{2}} = 1$

$\Rightarrow \frac{32}{5 a^{2}} + \frac{9}{b^{2}} = 1$ ……. (i)

From (i)

$\frac{6}{b^{2}} + \frac{9}{b^{2}} = 1 \Rightarrow b^{2} = 15 & a^{2} = 16$

$a^{2} + b^{2} = 15 + 16 = 31$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{25}$ coincide. Then the length of the latus rectum of the hyperbola is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Ellipse : $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$

Eccentricity = $\sqrt[]{1 - \frac{7}{10}} = \frac{3}{4}$

Foci $\equiv (\pm a e, 0)$ $\equiv (\pm 3, 0)$

Hyperbola : $\frac{x^{2}}{(\frac{144}{25})} - \frac{y^{2}}{(\frac{α}{25})} = 1$

Eccentricity = $\sqrt[]{1 + \frac{α}{144}} = \frac{1}{12} \sqrt[]{144 + α}$

$= 2 . \frac{\frac{81}{25}}{\frac{12}{5}} = \frac{27}{10}$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

The tangents at the points A(1, 3) and B(1, -1) on the parabola y² – 2x – 2y = 1 meet at the point P. Then the area (in unit²) of the triangle PAB is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

$y^{2} - 2 x - 2 y = 1$

$\Rightarrow {(y - 1)}^{2} = 2 (x + 1)$ …. (i)

Equation of tangent at A is 2x – y – 5 = 0 ………. (ii)

D is mid point of AB solving (ii) with y = 1 P (3, 1)

$∴ P D = 4, A D = 2$

$A r e a o f Δ A P D = \frac{1}{2} (P D) (A D) = 4$

$∴ A r e a o f Δ A P B = 8 s q . u n i t s$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

If the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ meets the line $\frac{x}{7} + \frac{y}{2 \sqrt[]{6}} = 1$ on the x-axis and the line $\frac{x}{7} - \frac{y}{2 \sqrt[]{6}} = 1$ on the y-axis, then the eccentricity of the ellipse is

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

Ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ passes through the points (7, 0) & (0, $- 2 \sqrt[]{6}$ )

$∴ a^{2} = 49 & b^{2} = 24$

$∴ e = \sqrt[]{1 - \frac{b^{2}}{a^{2}}} = \sqrt[]{1 - \frac{24}{49}} = \frac{5}{7}$

0 0

New answer posted

2 months ago

Conic Sections Class 11th

Let C be the centre of the circle $x^{2} + y^{2} - x + 2 y = \frac{11}{4}$ $^{}^{} \frac{}{}$ and P be a point on the circle. A line passes through the point C, makes an angle of $\frac{π}{4}$ $\frac{}{}$ with the line CP and intersects the circle at the points Q and R. Then the area of the triangle PQR (in unit2) is:

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

$x^{2} + y^{2} - x + 2 y = \frac{11}{4}$

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

or $Δ P Q R P R = Q R s i n 2 \geq \frac{1}{3}$

$= 4.6 s i n \frac{π}{8}$

$a s Δ P Q R = \frac{1}{2} P R * P Q$

$= 4 s i n \frac{π}{4} = \frac{4}{\sqrt[]{2}} = 2 \sqrt[]{2}$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
688k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Conic Sections

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience