If the line 3x – 4y = k touches the hyperbola x^2/5 - 4y^2/5 = 1, then find the value of k.

Question

If the line 3x – 4y = k touches the hyperbola \(\frac {x^2}{5} - \frac {4y^2}{5} = 1\), then find the value of k.

Answer 1

Given equation of the hyperbola is \(\frac {x^2}{5} - \frac {4y^2}{5} = 1\)

\(\frac {x^2}{5} - \frac {y^2}{\frac {5}{4}} = 1\)

Comparing this equation with \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\)

we get 

a² = 5, b² = 5/4

Given equation of line is 3x – 4y = k

y = 3/4 x - k/4

Comparing this equation with y = mx + c, we get

m = 3/4, c = - k/4

For the line y = mx + c to be a tangent to the hyperbola \(\frac {x^2}{a^2} - \frac {y^2}{b^2} = 1\),

we must have c² = a² m² – b²

⇒ k² = 25

⇒ k = ±5

Alternate method: 

Given equation of the hyperbola is

 \(\frac {x^2}{5} - \frac {4y^2}{5} = 1\) …….(i)

Given equation of the line is 3x – 4y = k

y = 3x-k/4

Substituting this value of y in (i), we get

⇒ 4x² – (9x² – 6kx + k² ) = 20 

⇒ 4x² – 9x² + 6kx – k² = 20 

⇒ -5x² + 6kx – k² = 20 

⇒ 5x² – 6kx + (k² + 20) = 0 …..(ii) 

Since, the given line touches the given hyperbola.

The quadratic equation (ii) in x has equal roots. 

(-6k)² – 4(5)(k² + 20) = 0 

⇒ 36k² – 20k² – 400 = 0 

⇒ 16k² = 400 

⇒ k² = 25 

⇒ k = ±5