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4 Maths -- Conic Section

Prove that the angle between tangents from (-2,3) to y2 = 8x are at right angles.

Prove that the angle between tangents from (-2,3) to y2 = 8x are at right angles.

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Here, the given equation of parabola is y2= 8x.

The equation of tangent to the parabola y2=8x is,   

               y= mx + 2/m

This tangent passes through the point (-2, 3)

So,       3 = -2m + 2/m

     or,  3m + 2m2 = 2

    or,   2m2 + 3m - 2 = 0

    or,   2m2 + (4 - 1)m -2 = 0

    or,   2m2 + 4m - m - 2 = 0

    or,   2m(m + 2) - 1(m+2) = 0

    or,   (m + 2) (2m - 1) = 0

    Either,                       Or,

       m = -2                  m = 1/2

Required angle is, 




Prove that the angle between tangents from (-2,3) to y2 = 8x are at right angles.

Here, the given equation of parabola is y2= 8x.

The equation of tangent to the parabola y2=8x is,   

               y= mx + 2/m

This tangent passes through the point (-2, 3)

So,       3 = -2m + 2/m

     or,  3m + 2m2 = 2

    or,   2m2 + 3m - 2 = 0

    or,   2m2 + (4 - 1)m -2 = 0

    or,   2m2 + 4m - m - 2 = 0

    or,   2m(m + 2) - 1(m+2) = 0

    or,   (m + 2) (2m - 1) = 0

    Either,                       Or,

       m = -2                  m = 1/2

Required angle is, 




More questions on Conic Section

Two ends of latus rectum are given then the maximum number of parabolas that can be drawn equal area. 0b. 1c. 2d. infinite

It is 2.

The maximum number of parabola that we can draw at the same time from the given extremities of Latus rectum is 2.

The length of the common chord of the parabola y2 =x and x2 =y isa. 1b. √2c. 2d. 1/2

The correct option is b.

Here,

The equations of given parabola are,

y2= x ..................(i)

and x2= y ..................(ii)

On solving eqs. (i) and (ii), we get,

x = 0, 1

and y = 0, 1

Since the points of intersection are (0, 0) and (1, 1).

Hence, length of common chord =(12+12)= 2

The locus of point which moves such that the tangent from which to a parabola y2 =4ax are at right angle. 

Let two tangents are drawn from point P.

Given equation of parabola is: y2=4ax.............(i)

We know that the locus of point P from which two perpendicular tangents are drawn to the parabola is the directrix of the parabola.

The standard equation of parabola (y - k)2 = 4p(x-h) has focus (x+h, k) and the directrix is (h - p).

From (i), we get,

h=0, k=0, p=1

Directrix of (i) is x= 0 - 1 = -1

Hence, the required locus is x= -1

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