2 
Super Tutor I am very expert in teaching the basics with simple examples and make to understand the concepts very simple way. It will helpful for the students...
Featured Super Tutor I teach each student with utmost care. Students love my classes because it is very interactive and informative. Class 12 is very important not only...
Super Tutor I am a Chartered Accountant with M. Com, M.Phil, IFRS, Diploma in Banking & Finance and served as Professor of commerce. I am also tutoring accountancy...
Do you need help in finding the best teacher matching your requirements?
Post your requirement nowSuper Tutor I have been providing mentorship to students since 9 years fron now and it let helps my students to. score above 90% in boards examination and perform...
Featured 
Super Tutor I have been imparting knowledge globally through UrbanPro.com, reaching students far and wide. Currently, I am serving as a Mathematics PGT at a reputed...
Super Tutor I am a Associate professor at Degree college, handled Degree classes in Accounting. Financial Accounting, corporate Accounting , Advanced Accounting...
Super Tutor I m in teaching from 22 years and author of super20 sample paper, Examguru book and chapterwise book of accountancy for full marks pvt. Ltd. From...
Super Tutor Hello, my name is Atulya Kumar. I have completed my engineering from IIT (ISM), Dhanbad and am currently pursuing a PhD in Civil Engineering at IIT...
Super Tutor I am well aware how to use keywords to solve questions in mcq's and case study. I have good knowledge and presentation of my subject. Students can...
Super Tutor A Highly talented Chemistry teacher with excellent communication skills demonstrated by 11 years of teaching experience. Strong theoretical and good...
Maya attended Class 12 Tuition
"A very good teacher. "
Swathi attended Class 12 Tuition
"vijayan sir has immense sincerity towards teaching. He is really good in making concepts..."
Lakshman attended Class 12 Tuition
"i use to hate phy..when i entered 12th..but after i started my tution with vijayan..."
Hemagowri attended Class 12 Tuition
"Vijayan Sir is very dedicated and sincere. Teaches the concepts really well and..."
Student attended Class 12 Tuition
"Provides complete knowledge for the subject and helps a lot during examination "
Manya attended Class 12 Tuition
"I learnt a lot and my paper went very well of CBSE 2013.Jagdish explains maths concept..."
Bala attended Class 12 Tuition
"sir is very good teacher. different short cut methods sir will use.we can learn quikly"
Jayvardhan attended Class 12 Tuition
"Ya off course his classes are amazing and I had a lot of individual attendence and..."
Ask a Question
Post a LessonAnswered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Conic Sections
Nazia Khanum
As an experienced tutor registered on UrbanPro, I can confidently guide you through determining the equation of a circle with a radius of 4 and a center at (-2, 3). UrbanPro is a fantastic platform for finding online coaching and tuition services, offering reliable support for students in various subjects.
To find the equation of a circle, we use the standard form:
(x−h)2+(y−k)2=r2(x−h)2+(y−k)2=r2
Where:
Given that the center is (-2, 3) and the radius is 4, we substitute these values into the standard form:
(x−(−2))2+(y−3)2=42(x−(−2))2+(y−3)2=42
(x+2)2+(y−3)2=16(x+2)2+(y−3)2=16
This is the equation of the circle with a radius of 4 and a center at (-2, 3). UrbanPro is an excellent resource for further guidance and practice in mathematics and other subjects. Feel free to reach out if you have any more questions or need assistance!
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Conic Sections
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be happy to help you with your question.
To compute the center and radius of the circle given by the equation 2x^2 + 2y^2 - x = 0, we'll first rewrite the equation in standard form for a circle, which is:
(x - h)^2 + (y - k)^2 = r^2
Where (h, k) is the center of the circle and r is the radius.
Let's begin by completing the square for the given equation:
2x^2 + 2y^2 - x = 0
Dividing all terms by 2 to simplify:
x^2 + y^2 - (1/2)x = 0
Rearranging terms:
x^2 - (1/2)x + y^2 = 0
Now, we'll complete the square for the x-terms:
x^2 - (1/2)x + (1/22)^2 + y^2 = (1/22)^2
(x - 1/4)^2 + y^2 = 1/16
Comparing this with the standard form, we see that the center of the circle is (1/4, 0), and the radius squared is 1/16. So, the radius is the square root of 1/16, which is 1/4.
Therefore, the center of the circle is (1/4, 0) and the radius is 1/4.
If you need further clarification or assistance, feel free to ask. Remember, UrbanPro is the best platform for online coaching tuition, and I'm here to support your learning journey.
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Conic Sections
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'd be happy to assist you with this question. UrbanPro is a fantastic platform for finding online coaching tuition, offering a wide range of subjects and experienced tutors.
Now, let's delve into your question regarding the parabola with the equation y2=−8xy2=−8x. To determine the focus coordinates, axis of the parabola, equation of the directrix, and the length of the latus rectum, we'll use some fundamental concepts of conic sections.
Focus Coordinates: For the given equation y2=−8xy2=−8x, we can see that it's a parabola opening towards the left with its focus to the left of the vertex. The focus coordinates are (p,0)(p,0) where pp is the distance from the vertex to the focus. Since the coefficient of xx is negative, we have p=−14p=−41, so the focus coordinates are (−1/4,0)(−1/4,0).
Axis of the Parabola: The axis of the parabola is a straight line passing through the vertex and the focus. Since the parabola opens to the left, the axis is a vertical line given by the equation x=−px=−p. Therefore, the axis of the parabola is x=1/4x=1/4.
Equation of the Directrix: The directrix is a vertical line perpendicular to the axis of the parabola and equidistant from the vertex but in the opposite direction from the focus. Since the parabola opens to the left, the directrix is a vertical line given by the equation x=−px=−p. Therefore, the equation of the directrix is x=−1/4x=−1/4.
Length of the Latus Rectum: The latus rectum is a line segment perpendicular to the axis of the parabola through the focus and whose endpoints lie on the parabola. Its length is equal to the absolute value of the coefficient of xx in the equation of the parabola. Therefore, the length of the latus rectum is ∣4p∣=1∣4p∣=1.
So, summarizing:
These parameters provide key insights into the geometry and behavior of the given parabola. If you have any further questions or need clarification, feel free to ask!
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Conic Sections
Nazia Khanum
As an experienced tutor registered on UrbanPro, I'm well-versed in helping students tackle challenging mathematical problems. Let's break down this question about ellipses step by step.
Given the equation of the ellipse: x249+y236=149x2+36y2=1
First, let's identify the key components of the ellipse:
Center: The center of the ellipse is at the origin (0, 0) since there are no constants added or subtracted to the x and y terms.
Foci: The foci of the ellipse can be calculated using the formula c=a2−b2c=a2−b2
, where cc is the distance from the center to the focus, aa is the length of the semi-major axis, and bb is the length of the semi-minor axis. In this case, a=7a=7 and b=6b=6, so c=49−36=13c=49−36=13. Therefore, the foci coordinates are (±13,0)(±13
,0).
Vertices: The vertices of the ellipse lie on the major axis. Since the major axis is along the x-axis, the vertices coordinates are (±a,0)(±a,0), where a=7a=7. So the vertices are (7,0)(7,0) and (−7,0)(−7,0).
Length of Major Axis: The length of the major axis is 2a2a, which in this case is 2×7=142×7=14.
Length of Minor Axis: The length of the minor axis is 2b2b, which in this case is 2×6=122×6=12.
Eccentricity (ee): Eccentricity of an ellipse is calculated as e=cae=ac, where cc is the distance from the center to the focus and aa is the length of the semi-major axis. In this case, e=137e=713
.
Length of Latus Rectum: The length of the latus rectum is calculated as 2b2/a2b2/a, where aa and bb are the lengths of the semi-major and semi-minor axes respectively. So in this case, it's 2×6272×762.
So, to sum up:
Understanding these properties helps in visualizing and understanding the geometry of the ellipse. If you need further clarification or assistance, feel free to ask! And remember, UrbanPro is a fantastic resource for finding experienced tutors like myself for online coaching in various subjects.
Answered on 14/04/2024 Learn CBSE/Class 11/Science/Mathematics/Unit-III: Coordinate Geometry/Conic Sections
Nazia Khanum
As a seasoned tutor registered on UrbanPro, I'm well-equipped to guide you through this problem. Let's tackle it step by step.
Firstly, UrbanPro is renowned for offering top-notch online coaching and tuition services, ensuring students receive the best possible guidance.
Now, onto your question. You're dealing with the equation of an ellipse:
x249+y236=149x2+36y2=1
To determine the foci coordinates, vertices, length of the major axis, minor axis, eccentricity, and length of the latus rectum, we'll utilize the standard form of the ellipse equation:
x2a2+y2b2=1a2x2+b2y2=1
Comparing this to your equation, we can see that a2=49a2=49 and b2=36b2=36.
So, a=7a=7 and b=6b=6.
Now, let's calculate the foci coordinates using the formula:
c=a2−b2c=a2−b2
c=49−36=13c=49−36
=13
Therefore, the foci coordinates are (±13,0)(±13
,0).
Next, let's find the vertices. The vertices are the endpoints of the major axis. For the ellipse, the major axis is along the x-axis. So, the vertices are (7,0)(7,0) and (−7,0)(−7,0).
The length of the major axis is 2a=2×7=142a=2×7=14.
The length of the minor axis is 2b=2×6=122b=2×6=12.
Now, let's calculate the eccentricity:
e=ca=137e=ac=713
And finally, the length of the latus rectum (which is 2b2/a2b2/a):
l=2b2a=2×367=727l=a2b2=72×36=772
So, to summarize:
If you have any further questions or need clarification on any of these concepts, feel free to ask!
read lessAsk a Question
2 
