AP EAPCET 2022 - 4th July Morning Shift | Hyperbola Question 7 | Mathematics

The locus of a variable point whose chord of contact w.r.t. the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$ subtends a right angle at the origin is

$$\frac{x^2}{4 a^2}-\frac{y^2}{4 b^2}=1$$

$$\left(\frac{x^2}{a^2}-\frac{y^2}{b^2}\right)=\frac{x^2}{a^4}+\frac{y^2}{b^4}$$

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

$$\frac{x^2}{a^4}+\frac{y^2}{b^4}=\frac{1}{a^2}-\frac{1}{b^2}$$

AP EAPCET 2021 - 20th August Morning Shift

MCQ (Single Correct Answer)

-0

If one focus of a hyperbola is $$(3,0)$$, the equation of its directrix is $$4 x-3 y-3=0$$ and its eccentricity $$e=5 / 4$$, then the coordinates of its vertex is

$$\left(\frac{3}{5}, \frac{11}{5}\right)$$

$$\left(\frac{11}{5}, \frac{3}{5}\right)$$

$$\left(\frac{7}{5}, \frac{4}{5}\right)$$

$$\left(\frac{4}{5}, \frac{7}{5}\right)$$

AP EAPCET 2021 - 19th August Morning Shift

MCQ (Single Correct Answer)

-0

The asymptotes of the hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$, with any tangent to the hyperbola form a triangle whose area is $$a^2 \tan (\alpha)$$. Then, its eccentricity equals

$$\sec (\alpha)$$

$$\operatorname{cosec}(\alpha)$$

$$\sec ^2(\alpha)$$

$$\operatorname{cosec}^2(\alpha)$$

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