Name: Conjugate Hyperbola
Uploaded: 2019-07-19
Duration: 12 min 19 s
Description: In this video, we will learn about the transverse axis and conjugate axis of any hyperbola with respect to another hyperbola. Watch and learn.

Question

A straight line is drawn parallel to the conjugate axis of a hyperbola to meet it and the conjugate hyperbola in the points

P

and

Q

respectively. Show that the tangents at

P

and

Q

meet on the curve

\frac{y^{4}}{b^{4}} (\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}) = \frac{4 x^{2}}{a^{2}}

and that the normals meet on the axis of

x

.

Accepted Answer

Let the equation of the hyperbola be:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 . . . (1)$
and the equation of its conjugate hyperbola be:
$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = - 1 . . . (2)$
Let $P$ be any point $(a \sec ϕ, b t a n ϕ)$ on equation $(1)$ .

Then the equation of the line parallel to the conjugate axis of $(1)$ i.e. $y - a x i s$ through point $P$ will be: $x = a \sec ϕ . . . (3)$

The line $(3)$ will cut the conjugate hyperbola $(2)$ at point $Q$ when $x = a \sec ϕ$ , and

Hence,
$y = b \sqrt{1 + \sec^{2} ϕ}$
So, the coordinate of $Q$ will be $(a \sec ϕ, b \sqrt{1 + \sec^{2} ϕ})$ .
Now, the equation of the tangent to $(1)$ at point $P$ is: $\frac{x}{a} - \frac{y}{b} \sin ϕ = \cos ϕ$
$\frac{x}{a} - \cos ϕ = \frac{y}{b} \sin ϕ . . . (4)$
And the equation of the tangent to $(2)$ at point $Q$ is: $\frac{y}{b} \sqrt{1 + \sec^{2} ϕ} - \frac{x}{a} \sec ϕ = 1$
$\frac{x}{a} + \cos ϕ = \frac{y}{b} \sqrt{1 + \cos^{2} ϕ} . . . (5)$

On squaring and adding equations $(4)$ and $(5)$ , we have: $\frac{2 x^{2}}{a^{2}} + 2 \cos^{2} ϕ = \frac{y^{2}}{b^{2}} [(1 + \cos^{2} ϕ) + \sin^{2} ϕ] = \frac{2 y^{2}}{b^{2}}$
$\cos^{2} ϕ = \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}$
Put the value of $\cos ϕ$ in equation $(5)$ , we get: $\frac{x}{a} + \sqrt{\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}} = \frac{y}{b} \sqrt{1 + \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}}$
Squaring, we get: $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}} + 2 . \frac{x}{a} \sqrt{\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}} = \frac{y^{2}}{b^{2}} (1 + \frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}})$
$\Rightarrow 2 . \frac{x}{a} \sqrt{\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}} = \frac{y^{2}}{b^{2}} (\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}})$
$\Rightarrow \frac{y^{2}}{b^{2}} \sqrt{\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}} = \frac{2 x}{a}$
Squaring, both sides, we get: $\frac{y^{4}}{b^{4}} (\frac{y^{2}}{b^{2}} - \frac{x^{2}}{a^{2}}) = \frac{4 x^{2}}{a^{2}}$
Again, the equations of the normal at point $P$ and $Q$ are: $a x \sin ϕ + b y + (a^{2} + b^{2}) \tan ϕ . . . (6)$

Now, $a x \sqrt{1 + \cos^{2} ϕ} + b y = (a^{2} + b^{2}) (\sqrt{(1 + \sec^{2} ϕ)}) . . . (7)$
After solving equations $(6)$ and $(7)$ , we have:
$x = \frac{a^{2} + b^{2}}{a} secϕ$ and $y = 0$ .
Hence, the two normals meet on the line $y = 0$ i.e., $x - a x i s$ .

Hence proved.

Important Questions on The Hyperbola

Learn and Prepare for any exam you want !

A straight line is drawn parallel to the conjugate axis of a hyperbola to meet it and the conjugate hyperbola in the points P and Q respectively. Show that the tangents at P and Q meet on the curve y4b4y2b2-x2a2=4x2a2 and that the normals meet on the axis of x.

Important Questions on The Hyperbola

Learn and Prepare for any exam you want !