Home
Class 12
MATHS
Find the slopes of the tangents to the p...

Find the slopes of the tangents to the parabola `y^2=8x` which are normal to the circle `x^2+y^2+6x+8y-24=0.`

Text Solution

Verified by Experts

The correct Answer is:
`(2pmsqrt(10))/(3)`
Any tangent to the parabola `y^(2)=8x" is "y=mx+2//m`, which is normal to the given circle.
Hence, the tangents must pass through the centre (-3, -4) of the circle.
Then, we have
`-4=-3m+(2)/(m)`
`or" "3m^(2)-4m-2=0`
`or" "m=(4pmsqrt(40))/(6)=(2pmsqrt(10))/(3)`
Promotional Banner

Topper's Solved these Questions

  • PARABOLA

    CENGAGE|Exercise Exercise 5.5|9 Videos

  • PARABOLA

    CENGAGE|Exercise Exercise 5.6|8 Videos

  • PARABOLA

    CENGAGE|Exercise Exercise 5.3|7 Videos

  • PAIR OF STRAIGHT LINES

    CENGAGE|Exercise Exercise (Numerical)|5 Videos

  • PERMUTATION AND COMBINATION

    CENGAGE|Exercise Question Bank|4 Videos

Similar Questions

Explore conceptually related problems

Find the equation of the tangent at t =2 to the parabola y^(2) = 8x .

Find the equations of the tangent and normal to the parabola y^(2)=8x at t=1/2

Find the equations of the tangent: to the parabola y^(2)=16x , parallel to 3x-2y+5=0

Find the equation of the tangent to the parabola x=y^2+3y+2 having slope 1.

Find the slope of the tangent to the curve y = x^(3) – x at x = 2.

The equation of the directrix of the parabola y^(2)=-8x is

Find the equations of the tangents to the circle x^2+y^2-6x+4y=12 which are parallel to the straight line 4x+3y+5=0

The vertex of the parabola x^(2)=8y-1 is :