The Focal Chord Of y2 = 16x is Tangent To (x – 6)2 + y2 = 2, Then The Possible Values Of The Slope Of This Chord, Are

Why Kaysons ?

Video lectures

Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.

Online Support

Practice over 30000+ questions starting from basic level to JEE advance level.

Live Doubt Clearing Session

Ask your doubts live everyday Join our live doubt clearing session conducted by our experts.

National Mock Tests

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

Organized Learning

Proper planning to complete syllabus is the key to get a decent rank in JEE.

Test Series/Daily assignments

Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.

SPEAK TO COUNSELLOR ? CLICK HERE

Question

The focal chord of y2 = 16x is tangent to (x – 6)2 + y2 = 2, then the possible values of the slope of this chord, are

Solution

Correct option is

1, –1

Focus (a, 0) is (4, 0). Any focal chord is

           y – 0 = m(x – 4)

or       mx – y – 4m = 0.

Apply the condition of tangency p = t with circle (6, 0), .

          

or     2m2 = m2 + 1  m2 = 1

.

SIMILAR QUESTIONS

Q1

The vertex of a parabola is the point (a, b) and latus-rectum is of length l. If the axis of the parabola is along the positive direction of y-axis. Then its equation is

Q2

Slope of common tangent to parabolas y2 = 4x and x2 = 8y is

Q3

If a focal chord with positive slope of the parabola y2 = 16xtouches the circle x2 + y2 – 12+ 34 = 0, then m is

Q4

If 2y = x + 24 is a tangent to parabola y2 = 24x, then its distance from parallel normal is

Q5

PQ is a focal chord of the parabola y2 = 4axO is the origin. Find the coordinates of the centroid, G, of triangle OPQ and hence find the locus of G as PQ varies.

Q6

Find the shortest distance between the circle x2 + y2 – 24y + 128 = 0 and the parabola y2 = 4x.

Q7

The equation of the directrix of the parabola y2 + 4y + 4x + 2 = 0 is  

Q8

If the line x – 1 = 0 is the directrix of the parabola y– ky + 8 = 0, then one of the of the value of k is

  

Q9

Equation of the parabola whose axis is y = x distance from origin to vertex is  and distance form origin to focus is , is (Focus and vertex lie in Ist quadrant) :

Q10

The curve described parametrically by x = t2 + t + 1, y = t2 – + 1 represents.