Question

The line $y = mx + 1$ is a tangent to the curve $y^2 = 4x$ if the value of $m$ is

• A. 1
• B. 2
• C. 3
• D. $\frac{1}{2}$

Answer 1

Answer: (A)

1

Answer 2

The correct answer is A:1
The equation of the tangent to the given curve is $y = mx + 1$. Now, substituting $y = mx + 1$ in $y^2 = 4x$, we get:
$(mx+1)^2=4x$
$m^2x^2+1+2mx-4x=0$
$m^2x^2+x(2m-4)+1=0 ...(i)$
Since a tangent touches the curve at one point, the roots of equation (i) must be equal. Therefore, we have:
Discriminant=0
$(2m-4)^2-4(m^2)(1)=0$
$=4m^2+16-16m-4m^2=0$
$16-16m=0$
$m=1$
Hence, the required value of m is 1.
The correct answer is A