Question

The line y=mx+1 touches y²=4x, if m=......,Select correct option from the given options.
(a) 0
(b) 1
(c) -1
(d) 2

Answer 1

HELLO DEAR,

The equation of the tangent to the given curve is y = mx + 1.

Now, substituting y = mx + 1 in y2 = 4x,

we get,

(mx + 1)² = 4x

(mx)² + 1 + 2mx = 4x

m²x² + 2x(m - 2) + 1 = 0------( 1 )

where, a = m² , b = (2m - 4) , c = 1

the line y = mx + 1 touches y² = 4x it is possible only when discriminant of ------- (1) will be zero.

so, Therefore,

discriminant (D) = b² - 4ac = 0

(2m - 4)² - 4*m²*1 = 0

4m² + 16 - 16m = 4m²

16 - 16m = 0

m = 1

Hence, the required value of m is 1.

The correct answer is b.

I HOPE ITS HELP YOU DEAR,.

THANKS

Answer 2

Hey there!

Q. The line y = mx + 1 touches y²= 4x, if m = ___.

(a) 0  

(b) 1  

(c) -1  

(d) 2

Answer : Option (b) 1

Solution :

Given, Line y = mx + 1 is tangent to the parabola ${\bold y^{2}}$ = 4x

Solving line y =mx + 1 with parabola ${\bold y^{2}}$, we have

⇒ $(mx + 1)^{2} = 4x$

⇒ $m^{2} x^{2} + 2mx + 1 = 4x$

⇒ $m^{2} x^{2} + x(2m - 4) + 1 = 0$

Since the line touches the parabola, the above equation must have equal roots.

Discriminant, D = 0

$(2m - 4)^{2} - 4m^{2} = 0$

$4m^{2} - 16m + 16 - 4m^{2} = 0$

16m = 16

m = $\frac{16}{16} = 1$

m = 1

Therefore, the value of m is 1.