Question

Solution

Correct option is

y = x + 2

A point on the curve xy = –1is not the form (t, –1/t). Now ∴ Equation of tangent to the curve xy = - 1 at (t, –1/t) is For this line to be tangent to the parabola y2 = 8x it should be of the form   Thus the required tangent is y = x + 2.

SIMILAR QUESTIONS

Q1

A dynamite blast blows a heavy rock starting up with a launch velocity to 160 m/sec. It reaches a height of s = 160t – 16t2 after t sec. The velocity of the rock when it is 256 m above ground on the way up is

Q2

The slope of the tangent to the curve represented by x = t2 + 3t – 8 and y = 2t2 – 2t – 5 at the point M (2, 1) is

Q3

The coordinates of the point P on the curve y2 = 2x3, the tangent at which is perpendicular to the line 4x – 3y + 2 = 0, are given by

Q4

The coordinates of points P(xy) lying in the first quadrant on the ellipsex2/8 + y2/18 = 1 so that the area of the triangle formed by the tangent at Pand the coordinate axes is the smallest, are given by

Q5

The points(s) on the curve y3 + 3x2 = 12y where the tangent is vertical is(are)

Q6

If ab > 0 then the minimum value of Q7

The curve y = ax3 + bx2 + cx + 8 touches x – axis at P(2, 0) and cuts they – axis at a point Q where its gradient is 3. The value of a, b, c are respectively

Q8

If the tangent at (1, 1) on y2 = x(2 – x)2 meets the curve again at P, is

Q9

The tangent to the curve At the point corresponding to is

Q10

The points of contact of the vertical tangents to x = 2 – 3 sinθ,  y = 3 + 2 cos θ are