The tangent drawn to the hyperbola x^ 2 16 - y^ 2 9 = 1, at point P...

The tangent drawn to the hyperbola

\dfrac{x^{2}}{16} - \dfrac{y^{2}}{9} = 1,

at point

P

in the first quadrant whose abscissa is

5,

meets the lines

3x - 4y = 0

and

3x + 4y = 0

Q

and

R

respectively. If

O

is the origin, then the area of triangle

OQR

is (in square units)

Options

A $6$
B $12$
C $3$
D $24$

Correct answer

B. $12$

Step-by-step solution

Let point

P

be \left(5, \beta\right)

since it lies on given hyperbola

\dfrac{25}{16} - \dfrac{{\beta}^{2}}{9} = 1 ⇒ {\beta}^{2} = \dfrac{81}{16}

<br />Hence,

\beta = \dfrac{9}{4}

(as

lies in first quadrant)<br />So, equation of tangent is

\dfrac{5x}{16} - \dfrac{9y}{4 ⋅ 9} = 1 ⇒ 5x - 4y = 16

<br />So, points~

Q and R

are \left(8, 6\right) and \left(2, - \dfrac{3}{2}\right)

Hence, the area of triangle

OQR

\dfrac{1}{2}\left(12 + 12\right) = 12

sq. units

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