JEE MATHEMATICS MCQ: HYPERBOLA

1 Easy

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Application +4 | -1

The locus of the poles of the chords of the hyperbola

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

which subtend a right angle at its centre is

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the locus of the poles can be found by homogenizing the hyperbola equation using the polar equation and then applying the condition for perpendicular lines at the center.

More Questions from HYPERBOLA

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Locus of the point of intersection of the tangents at the points with eccentric angles

\phi

and

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

on the hyperbola

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

is Let

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

be the point on the hyperbola

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

=36

, which is nearest to the line

3 x+2 y=1

. Then

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

is equal to Suppose an ellipse and a hyperbola have the same pair of foci on the

x

-axis with centres at the origin and they intersect at

(2,2)

. If the eccentricity of the ellipse is

\frac{1}{2}

, then the eccentricity of the hyperbola, is The eccentricity of the hyperbola

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

, is Let the hyperbola

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

pass through the point

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

. A parabola is drawn whose focus is same as the focus of

H

with positive abscissa and the directrix of the parabola passes through the other focus of

H

. If the length of the latus rectum of the parabola is

e

times the length of the latus rectum of

H

, where

e

is the eccentricity of

\mathrm{H}

, then which of the following points lies on the parabola? The equation of the conic with focus at

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

directrix along

x - y + 1 = 0

and with eccentricity

\sqrt 2