Find the range of c for which the line $y = m x + c$ touches the parabola $y^{2} = 8 (x + 2)$ .

Open in App

Solution

$y = m x + c y^{2} = 8 x + 16 o r (m x + c)^{2} = (8 x + 16) m^{2} x^{2} + 2 m c x + c^{2} = 8 x + 16 o r m^{2} x^{2} + (2 m c - 8) x + (c^{2} - 16) = 0 a s l i n e t o u c h e s p a r a b o l a, h e n c e o n l y o n e s o l u t i o n ∴ D = 0 (2 m c - 8)^{2} - 4. m^{2} . (c^{2} - 16) = 0 o r 4 (m c - 4)^{2} - m^{2} (c^{2} - 16) = 0 o r m^{2} c^{2} + 16 - 8 m c - m^{2} c^{2} + 16 m^{2} = 0 o r 16 m^{2} - 8 m c + 16 = 0 o r 2 m^{2} - m c + 2 = 0 a s m \in R ∴ D_{n} \geq 0 (- c)^{2} - 4.2.2 \geq 0 o r c^{2} \geq 16 c \in (- \infty, - 4] \cup [4, \infty)$

Suggest Corrections

Similar questions

m

c

y = m x + c

y^{2} = 8 (x + 2)

(- \infty, - k] \cup [k, \infty)

\frac{k}{2}

y = m x + c (m > 0)

y^{2} = 8 (x + 2)

c

y^{2} = 4 a (x + a)

c = m a + \frac{a}{m} .

y = m x + c

y^{2} = 4 a (x + a)

y = m x + c

y^{2} = 4 a (x + a)

Join BYJU'S Learning Program