MathematicsQ451–451 of 799 questions
Page 10 of 10 · English
451
MathematicsDifficultMCQAP EAMCET · 2022
The length of the latus rectum of the conic $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$ is
A
$\frac{1}{5}$
B
$\frac{2}{5}$
C
$\frac{3}{5}$
D
$\frac{4}{5}$
Solution
(D) The given equation is $25[(x-2)^2+(y-3)^2]=(3x-4y+7)^2$.
Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$.
This is in the form $SP^2 = PM^2$,where $S(2,3)$ is the focus and $3x-4y+7=0$ is the directrix.
Here,the distance from the focus to the directrix is $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}\right| = \left|\frac{6-12+7}{5}\right| = \frac{1}{5}$.
The length of the latus rectum of a parabola is $4a$.
Therefore,length $= 4 \times \frac{1}{5} = \frac{4}{5}$.
Dividing by $25$,we get $(x-2)^2+(y-3)^2 = \left(\frac{3x-4y+7}{5}\right)^2$.
This is in the form $SP^2 = PM^2$,where $S(2,3)$ is the focus and $3x-4y+7=0$ is the directrix.
Here,the distance from the focus to the directrix is $a = \left|\frac{3(2)-4(3)+7}{\sqrt{3^2+(-4)^2}}\right| = \left|\frac{6-12+7}{5}\right| = \frac{1}{5}$.
The length of the latus rectum of a parabola is $4a$.
Therefore,length $= 4 \times \frac{1}{5} = \frac{4}{5}$.
0 likesView Solution
Vedclass Products
For Students
Vedclass Test Series
Mock tests in real AP EAMCET style covering Mathematics with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D Mathematics papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Run live AP EAMCET mock exams with unlimited students, 360° analytics & white-label branding.
See DemoFrequently Asked Questions
How many Mathematics questions are in AP EAMCET 2022?
There are 799 Mathematics questions from the AP EAMCET 2022 paper on Vedclass, each with a detailed step-by-step solution in English.
Are AP EAMCET 2022 Mathematics solutions available in English?
Yes. All solutions on this page are in English. You can also switch to English or Hindi using the language buttons above the questions.
Can I practice AP EAMCET 2022 Mathematics as a timed test?
Yes. Use the Vedclass Test Series to attempt a full AP EAMCET mock test covering Mathematics with time limits and instant score analysis.
Can teachers create Mathematics papers from AP EAMCET previous year questions?
Yes. The Vedclass Exam Paper Generator lets teachers mix AP EAMCET Mathematics questions and generate Set A/B/C/D papers in minutes.
For Teachers & Institutes
Build a Custom Mathematics Paper
Pick AP EAMCET 2022 Mathematics questions, set difficulty, and generate Set A/B/C/D in 2 minutes.