JEE MATHEMATICS MCQ: HYPERBOLA

1 Easy

1 of 1

Application +4 | -1

\alpha+\beta=3 \pi

, then the chord joining the points

\alpha

and

\beta

for the hyperbola

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

passes through

View answer and explanation

Correct option: B

centre

Option B (centre) is correct because when

\alpha + \beta = 3\pi

, the equation of the chord simplifies to one that always passes through the origin (0,0), which is the center of the hyperbola.

More Questions from HYPERBOLA

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\phi

\frac{\pi}{2}-\phi

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

P\left(x_{0}, y_{0}\right)

3 x^{2}-4 y^{2}

=36

3 x+2 y=1

\sqrt{2}\left(y_{0}-x_{0}\right)

x

(2,2)

\frac{1}{2}

9 x^2-16 y^2+72 x-32 y-16=0

H: \frac{x^2}{a^2}-\frac{y^2}{b^2}=1

(2 \sqrt{2},-2 \sqrt{2})

H

H

e

H

e

\mathrm{H}

\left( {1, - 1} \right),

x - y + 1 = 0

\sqrt 2