JEE MATHEMATICS MCQ: HYPERBOLA

1 Easy

1 of 1

Application +4 | -1

The foci of a hyperbola coincide with the foci of the ellipse

\frac{x^2}{25}+\frac{y^2}{9}=1

. If the eccentricity of the hyperbola is 2 , then the equation of the tangent of this hyperbola passing through the point

(4,6)

View answer and explanation

Correct option: A

2 x-y-2=0

Option A (

2x - y - 2 = 0

) is correct because by finding the foci of the ellipse, then using the eccentricity of the hyperbola, we can determine the hyperbola's equation and subsequently find the tangent at the given point.

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