Equation of transverse and conjugate axis of hyperbola 16{x^2} - {y^2} + 64x + 4y + 44 = 0 are

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

medium

The equation of the transverse and conjugate axis of the hyperbola $16{x^2} - {y^2} + 64x + 4y + 44 = 0$ are

A

$x = 2,\;y + 2 = 0$

B

$x = 2,\;y = 2$

C

$y = 2,\;x + 2 = 0$

D

None of these

Solution

(c) ${(4x + 8)^2} – {(y – 2)^2} = – 44 + 64 – 4$

==> $\frac{{16{{(x + 2)}^2}}}{{16}} – \frac{{{{(y – 2)}^2}}}{{16}} = 1$

Transverse and conjugate axes are $y = 2$, $x = – 2$.

Standard 11

Mathematics

Share

Similar Questions

The graph of the conic $ x^2 – (y – 1)^2 = 1$ has one tangent line with positive slope that passes through the origin. the point of tangency being $(a, b). $ Then Length of the latus rectum of the conic is

normal

The distance between the directrices of a rectangular hyperbola is $10$ units, then distance between its foci is

easy

For the Hyperbola ${x^2}{\sec ^2}\theta – {y^2}cose{c^2}\theta = 1$ which of the following remains constant when $\theta $ varies $= ?$

medium

(AIEEE-2007)

If the length of the transverse and conjugate axes of a hyperbola be $8$ and $6$ respectively, then the difference focal distances of any point of the hyperbola will be

easy

Tangents are drawn to the hyperbola $4{x^2} – {y^2} = 36$ at the points $P$ and $Q.$ If these tangents intersect at the point $T(0,3)$ then the area (in sq. units) of $\Delta PTQ$ is :

hard

(JEE MAIN-2018)