If Pθ=x_1,(3√5)/2,0<θ<(π)/2 is a point on the hyperbola x^225-y^29=1, where θ is the parameter in its parametric form, then 2x_1+9sin^2θ=

Q101 / 160 · TS EAMCET 2022 (18 Jul Shift 2) · Mathematics

If $P (θ) = (x_{1}, \frac{3 \sqrt{5}}{2}), 0 < θ < \frac{π}{2}$ is a point on the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$ , where $θ$ is the parameter in its parametric form, then $2 x_{1} + 9 \sin^{2} θ =$

A
$8$
B
$10$
Correct answer
C
$20$
D
$34$

Solution

$\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$ is the given hyperbola.
Any point on the hyperbola is given by $(5 \sec θ, 3 \tan θ)$
Given point on the parabola is $P (θ) = (x_{1}, \frac{3 \sqrt{5}}{2}), 0 < θ < \frac{π}{2}$
i.e. $\tan θ = \frac{\sqrt{5}}{2}, \sec θ = \frac{3}{2}$
i.e. $x_{1} = \frac{15}{2}$
So, $2 x_{1} + 9 \sin^{2} θ = 15 + 9 {(\frac{\sqrt{5}}{3})}^{2} = 20$

Hence, the correct answer is option: 'C'.

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If Pθ=x1,352,0<θ<π2 is a point on the hyperbola x225-y29=1, where θ is the parameter in its parametric form, then 2x1+9sin2θ=

Solution

Relevant Questions

If $P (θ) = (x_{1}, \frac{3 \sqrt{5}}{2}), 0 < θ < \frac{π}{2}$ is a point on the hyperbola $\frac{x^{2}}{25} - \frac{y^{2}}{9} = 1$ , where $θ$ is the parameter in its parametric form, then $2 x_{1} + 9 \sin^{2} θ =$