Length of transverse axis of a hyperbola is 7 and it passes through point (5, -2) . equation of hype

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

medium

The length of the transverse axis of a hyperbola is $7$ and it passes through the point $(5, -2)$. The equation of the hyperbola is

A

$\frac{4}{{49}}{x^2} - \frac{{196}}{{51}}{y^2} = 1$

B

$\frac{{49}}{4}{x^2} - \frac{{51}}{{196}}{y^2} = 1$

C

$\frac{4}{{49}}{x^2} - \frac{{51}}{{196}}{y^2} = 1$

D

None of these

Solution

(c) Trick : $2a = 7$ or $a = \frac{7}{2}$

Also $(5, -2)$ satisfies it, so $\frac{4}{{49}}(25) – \frac{{51}}{{196}}(4) = 1$

and ${a^2} = \frac{{49}}{4}$

==> $a = \frac{7}{2}$.

Standard 11

Mathematics

Share

Similar Questions

The asymptote of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}}= 1$ form with any tangent to the hyperbola a triangle whose area is $a^2$ $\tan$ $ \lambda $ in magnitude then its eccentricity is :

normal

If $4{x^2} + p{y^2} = 45$ and ${x^2} – 4{y^2} = 5$ cut orthogonally, then the value of $p$ is

hard

An ellipse intersects the hyperbola $2 x^2-2 y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

$(A)$ Equation of ellipse is $x^2+2 y^2=2$

$(B)$ The foci of ellipse are $( \pm 1,0)$

$(C)$ Equation of ellipse is $x^2+2 y^2=4$

$(D)$ The foci of ellipse are $( \pm \sqrt{2}, 0)$

hard

(IIT-2009)

The straight line $x + y = \sqrt 2 p$ will touch the hyperbola $4{x^2} – 9{y^2} = 36$, if

easy

Area of the quadrilateral formed with the foci of the hyperbola $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = 1$ and $\frac{{{x^2}}}{{{a^2}}} – \frac{{{y^2}}}{{{b^2}}} = – 1$ is

normal