## Question

### Solution

Correct option is

x + y + 1 = 0

Tangent to  y2 = 4ax is … (1)

It passes through (1, –2). Then Also (1, –2) lies on y2 = 4ax a = 1

Put this in (2). m2 = –2m – 1 (m + 1)2 = 0 m = –1

And also   a = 1 y = –x – 1.

#### SIMILAR QUESTIONS

Q1

The equation of common tangent to the curves y2 = 8x and xy = –1 is

Q2

From the point (–1, 2) tangent lines are drawn to the parabola y2 = 4x, then the equation of chord of contact is

Q3

For the above problem, the area of triangle formed by chord of contact and the tangents is given by

Q4

A point moves on the parabola y2 = 4ax. Its distance from the focus is minimum for the following value(s) of x.

Q5

The line x – y + 2 = 0 touches the parabola y2 = 8x at the point

Q6

If t is the parameter for one end of a focal chord of the parabola y2 = 4ax, then its length is

Q7

The point on the parabola y2 = 8x at which the normal is inclined at 600 to the x-axis has the coordinates

Q8

The length of the latus rectum of the parabola 9x2 – 6x + 36y + 19 = 0 is

Q9

The equation of a circle passing through the vertex the extremities of the latus rectum of the parabola y2 = 8x  is

Q10

The equation of normal at the point to the parabola y2 = 4ax, is