A tangent to the hyperbola frac{{{x^2}}}{4} - | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

equation of the tangent at the point $'	heta '$ is

$\frac{{x\sec 	heta }}{a} - \frac{{y	an 	heta }}{b} = 1$

$ \Rightarrow P = \

Vedclass · Accepted Answer

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

Vedclass · Answer

equation of the tangent at the point $'	heta '$ is

$\frac{{x\sec 	heta }}{a} - \frac{{y	an 	heta }}{b} = 1$

$ \Rightarrow P = \left( {a\cos 	heta ,0} ight)\,\,\,\,Q = \left( {0, - b\cot 	heta } ight)$ 

Let $R$ be $\left( {h,k} ight) \Rightarrow h = a\cos 	heta ,k =  - b\cot 	heta $

$ \Rightarrow \frac{k}{h} = \frac{{ - b}}{{a\sin 	heta }} \Rightarrow \sin 	heta  = \frac{{ - bh}}{{ak}}$

$\cos 	heta  = \frac{h}{a}$

By squaring and adding,

$\frac{{{b^2}{h^2}}}{{{a^2}{k^2}}} + \frac{{{h^2}}}{{{a^2}}} = 1$

$ \Rightarrow \frac{{{b^2}}}{{{k^2}}} + 1 = \frac{{{a^2}}}{{{h^2}}}$

$ \Rightarrow \frac{{{a^2}}}{{{h^2}}} - \frac{{{b^2}}}{{{k^2}}} = 1$

Now, given $e{q^n}$ of hyperbola is $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$

$ \Rightarrow {a^2} = 4,{b^2} = 2$

$	herefore $ $R$ lies on $\frac{{{a^2}}}{{{x^2}}} - \frac{{{b^2}}}{{{y^2}}} = 1$ i.e., $\frac{4}{{{x^2}}} - \frac{2}{{{y^2}}} = 1$

A tangent to the hyperbola $\frac{{{x^2}}}{4} - \frac{{{y^2}}}{2} = 1$ meets $x-$ axis at $P$ and $y-$ axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on

Similar Questions

The equation of a tangent to the hyperbola $4x^2 -5y^2 = 20$ parallel to the line $x -y = 2$ is

The differential equation $\frac{{dx}}{{dy}}= \frac{{3y}}{{2x}}$ represents a family of hyperbolas (except when it represents a pair of lines) with eccentricity :

The equation of common tangents to the parabola ${y^2} = 8x$ and hyperbola $3{x^2} - {y^2} = 3$, is

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