If the line y = x + c touches the hyperbola x^ 2 9 - y^ 2 5 = 1 at the point P (h, k), then (h, k) can be equal to

Correct answer: B. ((9)/(2), (5)/(2)) Tangent at P(h, k) is (hx)/(9) - (ky)/(5) = 1 which is same as y = x + c (where c = ± 9(1) - 5 = ± 2 ) ⇒ (hx)/(9) - (ky)/(5) - 1 = 0 and x - y ± 2 = 0 are identical ⇒ ((h)/(9))/(1) = (- (k)/(5))/(- 1) = (- 1)/(± 2) ⇒ (h, k) = (- (9)/(2), - (5)/(2))or((9)/(2), (5)/(2))

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Hyperbola

NTA Abhyas 2020 · Mathematics · Hyperbola

If the line y = x + c touches the hyperbola x^ 2 9 - y^ 2 5 = 1 at...

If the line

y = x + c

touches the hyperbola

\dfrac{x^{2}}{9} - \dfrac{y^{2}}{5} = 1

at the point

P \left(h, k\right), then \left(h, k\right) can be equal to

Options

A $\left(\dfrac{9}{2}, - \dfrac{5}{2}\right)$
B $\left(\dfrac{9}{2}, \dfrac{5}{2}\right)$
C $\left(- \dfrac{5}{2}, - \dfrac{9}{2}\right)$
D $\left(\dfrac{5}{2}, - \dfrac{9}{2}\right)$

Correct answer

B. $\left(\dfrac{9}{2}, \dfrac{5}{2}\right)$

Step-by-step solution

Tangent at

P\left(h, k\right)

\dfrac{hx}{9} - \dfrac{ky}{5} = 1

which is same as

y = x + c

(where

c = \pm \sqrt{9\left(1\right) - 5} = \pm 2

)

⇒ \dfrac{hx}{9} - \dfrac{ky}{5} - 1 = 0

and

x - y \pm 2 = 0

are identical

⇒ \dfrac{\dfrac{h}{9}}{1} = \dfrac{- \dfrac{k}{5}}{- 1} = \dfrac{- 1}{\pm 2}

⇒ \left(h, k\right) = \left(- \dfrac{9}{2}, - \dfrac{5}{2}\right)or\left(\dfrac{9}{2}, \dfrac{5}{2}\right)

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