The equation of the tangent to the parabola y 2-8 x inclined at 30- to the x axis is A- 2 x y-14-0- 2 x-2 y-7-0- x-3 y-6-0D- 3 x-3 y-4-0

The correct option is C Slope of the tangent is (m) = Tan 30^∘ =1/√(3) It meets the parabola at y= 4√(3),x=6 Equation of the tangent having slope 1/√(3)and pa ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Slope Form of Tangent

The equation ...

Question

The equation of the tangent to the parabola $y^{2} = 8 x$ inclined at $30^{\circ}$ to the x axis is

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D
Slope of the tangent is (m) = $T a n 30^{\circ} = \frac{1}{\sqrt{3}}$
It meets the parabola at y= 4 $\sqrt{3}$ ,x=6
Equation of the tangent having slope $\frac{1}{\sqrt{3}}$ and passing through the above points is $x - \sqrt{3 y} + 6 = 0$

Suggest Corrections

Similar questions

Q. The equation of the tangent to the parabola

y^{2} = 16 x

inclined at an angle of

60^{\circ}

to the positive

x -

axis is

Q. x - axis is a tangent and y - axis is normal to a parabola whose focus is (2, 3)
The equation of tangent at vertex of parabola is

Q. The equation of tangent to the parabola

y^{2} = 6 x

at point

(6, 6)

Q. The asymptotes of a hyperbola have center at the point

(1, 2)

and are parallel to the lines

2 x + 3 y = 0

and

3 x + 2 y = 0

. If the hyperbola passes through the point

(5, 3)

, then its equation is

Q. Statement-I lf the line

x - 3 y + k = 0

touches the parabola

3 y^{2} = 4 x

then

k = 5

Statement - II: Equation of the tangent to

y^{2} = 8 x

inclined at an angle

30^{0}

to the axis is

x - \sqrt{3} y + 6 = 0

Join BYJU'S Learning Program

The equation of the tangent to the parabola y2=8x inclined at 30∘ to the x axis is

The correct option is D Slope of the tangent is (m) = Tan30∘=1√3 It meets the parabola at y= 4√3,x=6 Equation of the tangent having slope 1√3and passing through the above points is x−√3y+6=0

The equation of the tangent to the parabola $y^{2} = 8 x$ inclined at $30^{\circ}$ to the x axis is

The correct option is D
Slope of the tangent is (m) = $T a n 30^{\circ} = \frac{1}{\sqrt{3}}$
It meets the parabola at y= 4 $\sqrt{3}$ ,x=6
Equation of the tangent having slope $\frac{1}{\sqrt{3}}$ and passing through the above points is $x - \sqrt{3 y} + 6 = 0$