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Question

If the straight line $$y=mx+c$$ touches the hyperbola $$\displaystyle \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$ then prove that $$c^2 = a^2 m^2 - b^2$$


Solution

It is given that the line $$y=mx+c$$ touches the hyperbola. 
Consider the tangent of hyperbola in parametric form
$$\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$$
$$\therefore \dfrac{y\tan\theta}{b}=\dfrac{x\sec\theta}{a}-1$$
$$\therefore y=\dfrac{bx\sec\theta}{a\tan\theta}-\dfrac{b}{\tan\theta}$$
$$\therefore y=\dfrac{b}{a\sin\theta}x-\dfrac{b}{\tan\theta}$$
On comparing with $$y=mx+c$$, we get
$$m=\dfrac{b}{a\sin\theta}$$ and $$c=-\dfrac{b}{\tan\theta}$$
$$\implies a^2m^2-b^2=a^2\left(\dfrac{b}{a\sin\theta}\right)^2-b^2$$
$$\implies a^2m^2-b^2=\dfrac{b^2}{\sin^2\theta}-b^2$$
$$\implies a^2m^2-b^2=\dfrac{b^2(1-\sin^2\theta)}{\sin^2\theta}$$
$$\implies a^2m^2-b^2=\dfrac{b^2\cos^2\theta}{\sin^2\theta}$$
$$\implies a^2m^2-b^2=\dfrac{b^2}{\tan^2\theta}$$
$$\implies a^2m^2-b^2=c^2$$
Hence proved.

Maths

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