Hyperbola in conic sections

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$At the point of intersection of the rectangular hyperbola x y = c^{2} and the parabola y^{2} = 4 a x tangents to the$
$rectangular hyperbola and the parabola make angles θ and ϕ, respectively with x -axis, then$

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Solution

$Let (x_{1}, y_{1}) be point of intersection.$
$\begin{array}{ll} \Rightarrow y_{1}^{2} = 4 a x_{1}, x_{1} y_{1} = c^{2} \\ For y^{2} = 4 a x \end{array}$
$\begin{array}{l} m_{T_{(x_{1}, y_{1})}} = \frac{2 a}{y_{1}} = \tan ϕ \\ For x y = c^{2}, M_{T_{(x_{1}, y_{1})}} = - \frac{y_{1}}{x_{1}} = \tan θ \end{array}$
$\begin{array}{l} So \frac{\tan θ}{\tan ϕ} = \frac{- y_{1}^{2}}{2 a x_{1}} = - \frac{4 a x_{1}}{2 a x_{1}} = - 2 \\ So θ = \tan^{- 1} (- 2 \tan ϕ) \end{array}$

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Hyperbola in conic sections

Remember concepts with our Masterclasses.

At the point of intersection of the rectangular hyperbolaxy=c2 and the parabola y2=4ax tangents to the rectangular hyperbola and the parabola make angles θ and ϕ , respectively with x -axis, then

Let x1,y1 be point of intersection. ⇒ y12=4ax1,x1y1=c2 For y2=4axmTx1,y1=2ay1=tan⁡ϕ For xy=c2,MTx1,y1=−y1x1=tan⁡θ So tan⁡θtan⁡ϕ=−y122ax1=−4ax12ax1=−2 So θ=tan−1⁡(−2tan⁡ϕ)

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$At the point of intersection of the rectangular hyperbola x y = c^{2} and the parabola y^{2} = 4 a x tangents to the$
$rectangular hyperbola and the parabola make angles θ and ϕ, respectively with x -axis, then$