## Question

### Solution

Correct option is

C1 and C2 touch each other exactly at two points

Solving the two equations, we get

x2 + 4x – 6x + 1 = 0

⇒        x2 + 2x + 1 = 0

⇒        = 1 and y = ±2

So the two curve meet at two points (1, 2) and (1, –2). Equation of tangents at (1, 2) to C1is y(2) = 2 (x + 1)

⇒        y = x +1

and equation of the tangent at (1, 2) to C2 is x. 1 + y(2) – 3(+ 1) + 1 = 0  ⇒ y = x + 1

Showing that C1­ and C2 have a common tangent at the point (1, 2).

Similarly they have a common tangent

y = –(x + 1) at (1, –2)

Hence the two curve touch each other exactly at two points.

#### SIMILAR QUESTIONS

Q1

If F1 = (3, 0), F2 = (–3, 0) and P is any point on the curve 16x2 +25y2 = 400, then PF1 + PF2 equal

Q2

The locus of the points of intersection of the tangents at the extremities of the chords of the ellipse x2 + 2y2 = 6 which touch the ellipse x2 + 4y2 = 4 is

Q3

The normal at an end of a latus rectum of the ellipse passes through an end of the minor axis if

Q4

If an ellipse slides between two perpendicular straight line, then the locus of its center is

Q5

If the tangent at a point on the ellipse meets the auxillary circle in two points, the chords joining them subtends a right angle at the center; then the eccentricity of the ellipse is given by

Q6

The line passing through the extremity A of the major axis and extremity B of the minor axis of the ellipse. x2 + 9y2 = 9, meets the auxillary circle at the point M, then the area of the triangle with vertices at A, M and the origin is

Q7

The normal at a point P on the ellipse x2 + 4y2 = 16 meets the x-axis at Q, then locus of M intersects the latus rectums of the given ellipse at the points.

Q8

Let a and b be non-zero real numbers. Then the equation (ax2 + by2 + c)(x2 – 5xy + 6y2) = 0 represents

Q9

The pints of intersection of the two ellipse x2 + 2y2 – 6x – 12y + 23 = 0 and 4x2 + 2y2 – 20x – 12y + 35 = 0.