Math, asked by utkats, 17 days ago

if two tangents drawn from the point P to the parabola y2=12x be such that the slope of the tangent is double the other, then P lies on the curve

Answers

Answered by sukhmata22
4

Answer:

2y2=27x

Solution....

Given a parabola ,y2=12x

There is a point P outside the parabola with coordinates(h,k).Now the equation of a tangent to the parabola y2=4ax is given by

y=mx+ma;m is the slope of tangent

Here a=3

y=mx+m3

is the equation of any tangent to the parabola y2=12x.This tangent passes through point (h,k) hence

⇒m2h−km+3=0−(i)

So we can see that there are two possible values of ′m′ and hence we can have two tangents drawn to the parabola y2=12x from the point (h,k).

Also the slope of one is twice that of the other.

So say m1 & m2 are the roots of the quadratic equation

Then we have 

m1=2m2−(ii)

m1+m2=hk-(iii)

m1.m2=h3−(iv)

 2m22=h3−(v)

3m2=hk⇒m2=3hk−(vi)

Using (vi) we can reduce (v) as 9h22k2=h3

⇒2k2=27h

Replace k with y & h with x we have 

2y2=27x

Answered by amansharma264
9

EXPLANATION.

Two tangents drawn at a point ''P''.

To the parabola : y² = 12x.

Slope of tangent is double the other.

As we know that,

Concept of equation of tangent.

Equation of line : y = mx + c.

Equation of parabola : y² = 4ax.

Take two points on the parabola : (at², 2at).

Put the values of (at², 2at) in equation of line, we get.

⇒ 2at = mat² + c.

⇒ mat² - 2at + c = 0.

⇒ D = 0  Or b² - 4ac = 0.

⇒ (- 2a)² - 4(ma)(c) = 0.

⇒ 4a² - 4mac = 0.

⇒ a² - mac = 0.

⇒ a(a - mc) = 0.

⇒ c = a/m.

Put the value of c = a/m in equation of line, we get.

y = mx + a/m : Equation of tangents.

If tangents passes through point (h, k) then,

⇒ k = mh + a/m.

⇒ hm² - km + a = 0.

Using this concept in the equation, we get.

Equation of parabola : y² = 12x.

⇒ 4a = 12.

⇒ a = 3.

Put the value of a = 3 in equation of tangent, we get.

⇒ y = mx + 3/m.

⇒ k = mh + 3/m.

⇒ hm² - km + 3 = 0.

If m₁ and m₂ are the roots of the equation then,

Sum of the zeroes of the quadratic polynomial.

⇒ m₁ + m₂ = k/h. - - - - - (1).

Products of the zeroes of the quadratic polynomial.

⇒ m₁ m₂ = 3/h. - - - - - (2).

Slope of the tangent is double the other.

⇒ m₁ = 2m₂

Put the values of m₁ = 2m₂ in equation (), we get.

⇒ 2m₂ + m₂ = k/h.

⇒ 3m₂ = k/h.

⇒ m₂ = k/3h. - - - - - (3).

Put the values of m₁ = 2m₂ in equation (2), we get.

⇒ (2m₂) x (m₂) = 3/h.

⇒ (2m₂)² = 3/h. - - - - - (4).

Put the values of equation (3) in equation (4), we get.

⇒ 2(k/3h)² = 3/h.

⇒ 2(k²/9h²) = 3/h.

⇒ 2k²/9h² = 3/h.

⇒ 2k²h = 27h².

⇒ 2k² = 27h.

⇒ 2y² = 27x.

P lies on the curves : 2y² = 27x.

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