The straight line y=m x+c(m0) touches the parabola y2=8(x+2). Then the minimum value taken by c is

Question

The straight line y=m x+c(m>0) touches the parabola y2=8(x+2). Then the minimum value taken by c is (A) 4 (B) 8 (C) 12 (D) 6

Tardigrade Admin · Accepted Answer

Tangent to y2=8(x+2) having slope ' m ' is y=m(x+2)+(2/m) ∴ c=2 m+(2/m) ⇒ (c/2)=(m+(1/m)) ∵ m+(1/m) ≥ 2 ⇒ (c/2) ≥ 2 ⇒ c ≥ 4 Thus, the minimum value of c is 4 .

Q. The straight line $y = m x + c (m > 0)$ touches the parabola $y^{2} = 8 (x + 2)$ . Then the minimum value taken by $c$ is

4

8

12

6

Tangent to $y^{2} = 8 (x + 2)$ having slope ' $m$ ' is
$y = m (x + 2) + \frac{2}{m}$
$∴ c = 2 m + \frac{2}{m}$
$\Rightarrow \frac{c}{2} = (m + \frac{1}{m})$
$∵ m + \frac{1}{m} \geq 2$
$\Rightarrow \frac{c}{2} \geq 2 \Rightarrow c \geq 4$
Thus, the minimum value of $c$ is $4$ .

Q. The straight line y=mx+c(m>0) touches the parabola y2=8(x+2). Then the minimum value taken by c is

4

8

12

6

Tangent to y2=8(x+2) having slope ' m ' is y=m(x+2)+m2​ ∴c=2m+m2​ ⇒2c​=(m+m1​) ∵m+m1​≥2 ⇒2c​≥2⇒c≥4 Thus, the minimum value of c is 4 .

Q. The straight line $y = m x + c (m > 0)$ touches the parabola $y^{2} = 8 (x + 2)$ . Then the minimum value taken by $c$ is

Tangent to $y^{2} = 8 (x + 2)$ having slope ' $m$ ' is
$y = m (x + 2) + \frac{2}{m}$
$∴ c = 2 m + \frac{2}{m}$
$\Rightarrow \frac{c}{2} = (m + \frac{1}{m})$
$∵ m + \frac{1}{m} \geq 2$
$\Rightarrow \frac{c}{2} \geq 2 \Rightarrow c \geq 4$
Thus, the minimum value of $c$ is $4$ .