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

Correct answer: (B) 9/2,5/2. Tangent at Ph,k is hx/9-ky/5=1 which is same as y=x+c (where c=±√(91-5)=±2 ) ⇒hx/9-ky/5-1=0 and x-y±2=0 are identical ⇒h/91=(-k)/5-1=(-1)/(±2) ⇒h,k=-9/2,-5/2 or 9/2,5/2

Q30 / 40 · NTA Abhyas (JEE Main) 2020 · Mathematics · Hyperbola

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

A
$(\frac{9}{2}, - \frac{5}{2})$
Correct answer
B
$(\frac{9}{2}, \frac{5}{2})$
C
$(- \frac{5}{2}, - \frac{9}{2})$
D
$(\frac{5}{2}, - \frac{9}{2})$

Solution

Tangent at

P (h, k)

\frac{h x}{9} - \frac{k y}{5} = 1

which is same as

y = x + c

(where

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

)

\Rightarrow \frac{h x}{9} - \frac{k y}{5} - 1 = 0

and

x - y \pm 2 = 0

are identical

\Rightarrow \frac{\frac{h}{9}}{1} = \frac{- \frac{k}{5}}{- 1} = \frac{- 1}{\pm 2}

\Rightarrow (h, k) = (- \frac{9}{2}, - \frac{5}{2}) o r (\frac{9}{2}, \frac{5}{2})

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

If the line y=x+c touches the hyperbola x29-y25=1 at the point Ph,k, then h,k can be equal to

Solution

More Hyperbola Questions

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