Класс 11 RD Sharma Solutions — Глава 25 Парабола — Упражнение 25.1 | Набор 2

Вопрос 11. Найдите уравнение параболы с фокусом в точке (5, 2) и вершиной в точке (3, 2).

Решение:

Given that, the vertex of the parabola is (3, 2) and the focus of the parabola is (5, 2)

So, the slope of the axis of the parabola = 0

The slope of the directrix is not defined.

Let us assume the directrix intersect the axis at point K (r, s).  

So, (r + 5)/2 = 3, (s + 2)/2 = 2

=> r = 1, s = 2

Hence, the equation of the directrix is x – 1 = 0 or x = 1.

Now, let us assume P (x, y) be any point on the parabola whose focus is S (5, 2).

And the equation of directrix is x =1.

So, the first we draw PM perpendicular to x  = 1.

Then

SP = PM

SP² = PM²

(x – 5)² + (y – 2)² = 

x² + 25 – 10x + y² + 4 – 4y = x² + 1 – 2x

25 – 10x + y² + 4 – 4y – 1 + 2x = 0

=> y² – 4y – 8x + 28 = 0

Hence, the equation of the parabola is y² – 4y – 8x + 28 = 0

Вопрос 12. Канат равномерно нагруженного висячего моста висит в виде параболы. Горизонтальная проезжая часть длиной 100 м поддерживается вертикальными тросами, прикрепленными к кабелю, самый длинный из которых составляет 30 м, а самый короткий — 6 м. Найдите длину опорного троса, прикрепленного к проезжей части на расстоянии 18 м от середины.

Решение:

Given that the suspension cable forms a parabola with the vertex at (0, 6).

So, let us assume that the equation of the parabola formed by the suspension cable be,  

(x – 0)² = 4a (y – 6)

And it passes through point P (−50, 30) and Q (50, 30).  

So, 2500 = 4a (30 – 6)

=> 4a = 2500/24

By putting the value of 4a in the given equation, we get

x² = (2500/24) (y – 6)

Now, let the coordinates be (18, l) and it lies on the parabola (2).  

So, 18² = (2500/24) (l – 6)

=> 324 = (2500/24) (l – 6)

=> l = 9.11 m

Therefore, 9.11 m is the supporting wire attached to the roadway 18 m from the middle.

Вопрос 13. Найдите уравнения прямых, соединяющих вершину параболы y ² = 6x с точкой на ней, имеющие абсциссу 24.

Решение:

Let us assume A and B be points on the parabola y² = 6x.

Now, OA, OB be the lines joining the vertex O to the points A and B whose abscissa is 24.  

Now,

=> y² = 6 × 24 

=> y² = 144

=> y = ± 12

So, the coordinates of the points A is (24, 12) and B is (24, –12).

Hence, the equation of lines are 

=> 

=> ±2y = x

Вопрос 14. Найдите координаты точек на параболе y ² = 8x, фокусное расстояние которых равно 4.

Решение:

Given that, the parabola is y² = 8x 

⇒ y² = 4(2)x

Now, on comparing it with the general equation of parabola y² = 4ax, we will get a = 2.

Let us assume that the required point be (x₁, y₁).

Also, given that focal distance is 4

=> x₁ + a = 4

=> x₁ + 2 = 4

=> x₁ = 2

Now, this point satisfy the equation of parabola,

So, (y₁)² = 8(2) = 16

=> y₁² = 16

=> y₁ = ± 4

Hence, the coordinates of the points are (2, 4) and (2, −4).

Вопрос 15. Найдите длину отрезка, соединяющего вершину параболы y ² = 4ax и точку на параболе, в которой отрезок образует угол θ с осью x.

Решение:

Let us assume that the coordinates of the point on the parabola is B (x₁, y₁) and BO be the line segment in the parabola.

Now, in triangle AOB, 

cos θ = AO/OB and sin θ = AB/OB

=> cos θ = x₁/OB  and sin θ = y₁/OB

x₁ = OB cos θ and y₁ = OB sin θ

Now, the curve is passing through point (x₁, y₁)

So, (y₁)² = 4a(x₁)

=> (OB sin θ)² = 4a (OB cos θ)  

=> (OB)² sin2 θ = 4a OB cos θ

=> OB = 4a cos θ/sin² θ

=> OB = 4a cosec θ cot θ

Therefore, the required length is 4a cosec θ cot θ.

Вопрос 16. Если точки (0, 4) и (0, 2) являются соответственно вершиной и фокусом параболы, то найдите уравнение параболы.

Решение:

Given that the vertex of the parabola is (0, 4) and the focus of the parabola is (0, 2)

From the points we conclude that the vertex and focus lie on y-axis, so y-axis is the axis of the parabola.

Now, if the directrix meets the axis of the parabola at point Z, then AZ = AF = 2.

OZ = OF + AZ + FA 

= 2 + 2 + 2 

= 6

So, the equation of the directrix is y = 6.

i.e., y − 6 = 0

Let us assume P(x, y) be any point in the plane of the focus and directrix. 

And MP be the perpendicular distance from P to the directrix, then P lies on parabola if FP = MP.

x² + y² – 4y + 4 = y² – 12y + 36

=> x² + 8y = 32

Hence, the equation of the parabola is x² + 8y = 32

Вопрос 17. Если прямая y = mx + 1 касается параболы y ² = 4x, то найти значение m.

Решение:

Given that the equation of the is y² = 4x.

On substituting the value of y = mx + 1 in the equation of parabola, we get

(mx + 1)² = 4x

⇒ m²x² + 2mx + 1 = 4x

⇒ m²x² + (2m − 4)x + 1 = 0           

As we know that a tangent touches the curve at a point, so the roots of the equation must be equal.

So, D = 0

=> (2m − 4)² − 4m² = 0

=> 4m² −16m + 16 − 4m² = 0

=> m = 1

Therefore, the value of m is 1.

Вопрос 18. Найдите расстояние между вершиной и фокусом параболы y ² + 6y + 2x + 5 = 0.

Решение:

Given that the equation of the parabola is y² + 6y + 2x + 5 = 0

(y + 3)² + 2x – 4 = 0

(y + 3)² = -2 (x – 2)

Let us assume Y = y + 3 and X = x – 2.

Now we get

Y² = – 2X

On putting 4a = 2, we get

=> a = 1/2

Focus = (X = -1/2, Y = 0) = (x = 3/2, y = – 3)

Vertex = (X = 0, Y = 0) = (x = 2, y = -3)

So, 

Focus = (3/2, -3)

Vertex = (2, -3)

Now, we find the distance between the vertex and the focus is,

D = 

= 

= 1/2 units

Therefore, the required distance is 1/2 units.

Вопрос 19. Найдите уравнение направляющей параболы x ² − 4x − 8y + 12 = 0.

Решение:

Given that the equation of the parabola is x² − 4x − 8y + 12 = 0  

(x – 2)² – 4 – 8y + 12 = 0

(x – 2)² = 8 (y – 1) 

Let us assume Y = y − 1 and X = x – 2. 

Now we get

X² = 8Y

Now, on comparing with x² = 4ay, we get

=> a = 2

So, Directrix = Y = −a

=> y − 1 = −a

=> y = −a + 1

=> y = −2 + 1

=> y = −1  

Therefore, the required equation of the directrix is y = -1.

Вопрос 20. Напишите уравнение параболы с фокусом (0, 0) и направляющей x + y − 4 = 0.

Решение:

Given that the focus (0, 0) and directrix x + y − 4 = 0 of the parabola

Now, let us assume P (x, y) be any point on the parabola with the given focus and directrix

So, first we draw PM perpendicular to x + y = 4.

Then, 

SP = PM

SP² = PM²

(x – 0)² + (y – 0)² = 

x² + y² = 

2x² + 2y² = x² + y² + 16 + 2xy – 8y – 8x

=> x² + y² – 2xy + 8x + 8y – 16 = 0

Hence, the equation of the parabola is x² + y² – 2xy + 8x + 8y – 16 = 0