VITEEE 2024 | Ellipse Question 1 | Mathematics

Consider the ellipse $\frac{x^2}{\cos ^2 \alpha}+\frac{y^2}{\sin ^2 \alpha}=1$, where $\alpha \in\left(0, \frac{\pi}{4}\right)$. Then, locus of point of intersection of one of the directrix and tangent at upper end of minor axis is

$x \sqrt{1-y^2}=\left(1-2 y^2\right)$

$x \sqrt{1-2 y^2}=\left(1-y^2\right)^2$

$x^2\left(1-y^2\right)=\left(1-2 y^2\right)$

$x^2\left(1-2 y^2\right)=\left(1-y^2\right)^2$

VITEEE 2022

MCQ (Single Correct Answer)

-0

If a man running around a race-course notes that the sum of the distances of two flag-posts from him is always $$10 \mathrm{~m}$$ and the distance between the flag-posts is $$8 \mathrm{~m}$$, then the area of the path he encloses in square metres is

12$$\pi$$

13$$\pi$$

14$$\pi$$

15$$\pi$$

VITEEE 2021

MCQ (Single Correct Answer)

-0

The focal distance of the point $$(x, y)$$ from the ellipse $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$$ is

$$a \pm \sqrt{1-\frac{b^2}{a^2}} y$$

$$b \pm \sqrt{1-\frac{a^2}{b^2}} y$$

$$a \pm \sqrt{1-\frac{b^2}{a^2}} x$$

$$b \pm \sqrt{1-\frac{a^2}{b^2}} x$$

VITEEE 2021

MCQ (Single Correct Answer)

-0

The eccentricity of the ellipse $$25 x^2+9 y^2-150 x-90 y+225=0$$ is

$$\frac{4}{5}$$

$$\frac{3}{5}$$

$$\frac{4}{15}$$

$$\frac{9}{5}$$

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Aptitude

Coding and Decoding Number Series Syllogism Data Sufficiency Data Interpretation Analogy Classification Series Completion Blood Relations Logical Deduction

English

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