Question

Equation of a common tangent to the curves y2 = 8x and xy = –1 is 

Solution

Correct option is

x + 2

Equation of a tangent at (at2, 2at) to y2 = 8x is ty = x + atwhere 4a = 8 i.e., = 1.

⇒ ty = x + at2 which intersects the curve xy = –1 at the points given by  or x2 + 2t2x + t = 0 and will be a tangent to the curve if the roots of this quadratic equation are equal, for which 4t4 – 4t = 0 ⇒ t = 0 or t = 1 and an equation of a common tangent is x + 2.

SIMILAR QUESTIONS

Q1

If P, Q, R are three points on a parabola y2 = 4ax whose ordinates are in geometrical progression, then the tangents at and R meet on

Q2

If Land L2 are the length of the segments of any focal chord of the parabola y2 = x, then  is equal to

Q3

The tangents at three points A, B, C on the parabola y2 = 4x, taken in pairs intersect at the points Pand R. If  be the areas of the triangles ABC and PQR respectively, then 

Q4

The locus of the mid-point of the line segment joining the focus to a moving point on the parabola y2 = 4ax is another parabola with directrix

Q5

The tangent at the point P(x1y1) to the parabola y2 = 4ax meets the parabola y2 = 4a(x + b) at Q and R, the coordinates of the mid-point of QR are  

Q6

Consider a parabola y2 = 4ax, the length of focal chord is l and the length of the perpendicular from vertex to the chord is pthen

Q7

Tangent are drawn to a parabola from a point T. If P, Q are the points of constant then perpendicular distance from P, T and upon the tangent at the vertex of the parabola are in.

Q8

Chord of the parabola  which subtend right angle at vertex pass through