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  • If the tangent at the point <img alt="\mathrm{(2 \sec \phi, 3 \tan \phi)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%282%20%5Csec%20%5Cphi%2C%203%20%5Ctan%20%5Cphi%29%7

If the tangent at the point \mathrm{(2 \sec \phi, 3 \tan \phi)} on the hyperbola \mathrm{\frac{x^2}{4}-\frac{y^2}{9}=1} is parallel to \mathrm{3 x-y+4=0}, then the value of \mathrm{\phi } is 

Option: 1

45^{\circ}


Option: 2

60^{\circ}


Option: 3

30^{\circ}


Option: 4

75^{\circ}


Answers (1)

best_answer

Here \mathrm{x=2 \sec \phi \text { and } y=3 \tan \phi}
Differentiating w.r.t. \mathrm{\varphi }
\mathrm{\frac{d x}{d \phi}=2 \sec \phi \tan \phi\ \ and \quad \frac{d y}{d \phi}=3 \sec ^2 \phi}
\mathrm{\therefore } Gradient of tangent \mathrm{\frac{d y}{d x}=\frac{d y / d \phi}{d x / d \phi}=\frac{3 \sec ^2 \phi}{2 \sec \phi \tan \phi} ; \quad \therefore \quad \frac{d y}{d x}=\frac{3}{2} \operatorname{cose} \phi\ \ \ \ \ .........(i)}
But tangent is parallel to \mathrm{3 x-y+4=0\ ;\ \ \therefore} Gradient \mathrm{m=3\ \ \ ............(ii)}
From \mathrm{(i)} and \mathrm{(ii)}, \mathrm{\frac{3}{2} \operatorname{cose} \phi=3 \Rightarrow \operatorname{cose} \phi=2,\ \ \therefore \phi=30^{\circ}}

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