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 line through the origin meets the circle

x^{2} + y^{2} = a^{2}

at

P

and the hyperbola

x^{2} - y^{2} = a^{2}

at

Q

. Prove that the locus of the point of intersection of tangent at

P

to the circle with the tangent at

Q

to the hyperbola is the curve.

Accepted Answer

Let, the line $y = x \tan θ$ meet the circle $x^{2} + y^{2} = a^{2}$ at $P$ and a hyperbola $x^{2} - y^{2} = a^{2}$ at $Q$ .

Here, $P (a \cos θ, a \sin θ)$ .

For the point $Q$

$\Rightarrow x^{2} - {(x \tan θ)}^{2} = a^{2}$

$\Rightarrow x^{2} = \frac{a^{2}}{1 - \tan^{2} θ}$

$\Rightarrow Q = (\frac{a}{\sqrt{1 - \tan^{2} θ}}, \frac{a \tan θ}{\sqrt{1 - \tan^{2} θ}})$

Now, the equation of tangent of circle $x^{2} + y^{2} = a^{2}$ at $P (a \cos θ, a \sin θ)$ is $x \cos θ + y \sin θ = a . . . (i)$

And the equation of tangent of hyperbola $x^{2} - y^{2} = a^{2}$ at $Q (\frac{a}{\sqrt{1 - \tan^{2} θ}}, \frac{a \tan θ}{\sqrt{1 - \tan^{2} θ}})$ is $x - y \tan θ = a \sqrt{1 - \tan^{2} θ}$ .

$\Rightarrow x \cos θ - y \sin θ = a \sqrt{\cos^{2} θ - \sin^{2} θ} . . . . (i i)$

Let, $(h, k)$ be the point of intersection of tangents of given circle and hyperbola.

$\Rightarrow h = \frac{a + a \sqrt{\cos^{2} θ - \sin^{2} θ}}{2 \cos θ}, k = \frac{a - a \sqrt{\cos^{2} θ - \sin^{2} θ}}{2 \sin θ}$

$\Rightarrow h = \frac{a + a \sqrt{\cos^{2} θ - \sin^{2} θ}}{2 \cos θ}, k = \frac{a - a \sqrt{\cos^{2} θ - \sin^{2} θ}}{2 \sin θ}, h k = \frac{a^{2} (1 - \cos^{2} θ + \sin^{2} θ)}{4 \sin θ \cos θ}$

$\Rightarrow \frac{2 h}{a} = \sec θ + \sqrt{1 - \tan^{2} θ}, h k = \frac{a^{2}}{2} \tan θ$

$\Rightarrow \frac{2 h}{a} = \pm \sqrt{1 + {(\frac{2 h k}{a^{2}})}^{2}} + \sqrt{1 - {(\frac{2 h k}{a^{2}})}^{2}}$

$\Rightarrow \frac{2 h}{a} - \sqrt{1 - \frac{4 h^{2} k^{2}}{a^{4}}} = \pm \sqrt{1 + \frac{4 h^{2} k^{2}}{a^{4}}}$

$\Rightarrow {(\frac{2 h}{a} - \sqrt{1 - \frac{4 h^{2} k^{2}}{a^{4}}})}^{2} = {(\pm \sqrt{1 + \frac{4 h^{2} k^{2}}{a^{4}}})}^{2}$

$\Rightarrow \frac{4 h^{2}}{a^{4}} + 1 - \frac{4 h^{2} k^{2}}{a^{4}} - \frac{4 h}{a} \sqrt{1 - \frac{4 h^{2} k^{2}}{a^{4}}} = 1 + \frac{4 h^{2} k^{2}}{a^{4}}$

$\Rightarrow h^{2} - a h \sqrt{a^{4} - 4 h^{2} k^{2}} = 2 h^{2} k^{2}$

$\Rightarrow {(h - 2 h k^{2})}^{2} = {(a \sqrt{a^{4} - 4 h^{2} k^{2}})}^{2}$

$\Rightarrow h^{2} + 4 h^{2} k^{4} - 4 h^{2} k^{2} = a^{6} - 4 a^{2} h^{2} k^{2}$

$\Rightarrow h^{2} + 4 h^{2} k^{4} + 4 (a^{2} - 1) h^{2} k^{2} = a^{6}$

Replace $h$ by $x$ and $k$ by $y$ .

$\Rightarrow x^{2} + 4 x^{2} y^{4} + 4 (a^{2} - 1) x^{2} y^{2} = a^{6}$

A line through the origin meets the circle $x^{2} + y^{2} = a^{2}$ at $P$ and the hyperbola $x^{2} - y^{2} = a^{2}$ at $Q$ . Prove that the locus of the point of intersection of tangent at $P$ to the circle with the tangent at $Q$ to the hyperbola is the curve.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

A line through the origin meets the circle x2+y2=a2 at P and the hyperbola x2-y2=a2 at Q. Prove that the locus of the point of intersection of tangent at P to the circle with the tangent at Q to the hyperbola is the curve.

Important Questions on Hyperbola

Learn and Prepare for any exam you want !

A line through the origin meets the circle $x^{2} + y^{2} = a^{2}$ at $P$ and the hyperbola $x^{2} - y^{2} = a^{2}$ at $Q$ . Prove that the locus of the point of intersection of tangent at $P$ to the circle with the tangent at $Q$ to the hyperbola is the curve.