Students must practice these TS Inter 2nd Year Maths 2B Important Questions Chapter 3 Parabola to help strengthen their preparations for exams.

## TS Inter 2nd Year Maths 2B Parabola Important Questions

Very Short Answer Type Questions

Question 1.

Find the coordinates of the vertex and focus and the equations of the directrix and axis of the following parabolas.

(i) y^{2} = 16x

Comparing with y^{2} = 4ax, we have 4a = 16

a = 4

Vertex = (0, 0)

Focus = (a, 0) = (4, 0)

Equation of clirectrix is x + a = 0 ⇒ x + 4 = 0

Equation of axis is y = 0.

(ii) x^{2} =-4y

Comparing with x^{2} = – 4ay, we have 4a = 4 ⇒ a = I

∴ Vertex = (0. 0)

Focus =(0,-a)=(0,-1)

Equation of directrix is y-a=0=y-1 =0

Equation of axis is x – 0.

(iii) 3x- 9x + 5y – 2 = 0

Given equation is 3x^{2} – 9x – 5y +2

= 3(x^{2} – 3x) =- \(5\left(\mathrm{y}-\frac{2}{5}\right)\)

(i) Vertex (h, k) \(\left(\frac{3}{2}, \frac{7}{4}\right)\)

(ii) Focus=(h,k,-a) = \(\left(\frac{3}{2}, \frac{7}{4}-\frac{5}{12}\right)=\left(\frac{3}{2}, \frac{4}{3}\right)\)

(iii) Equation of directrix is y – k = 0

⇒ 6y – 13= 0

(iv) Equation of axis x – h = 0

= x = \(\frac{3}{2}\) – 0 ⇒ 2x – 3 = 0

(iv) y^{2} – x + 4y + 5 = 0

The given equation can be written as

y^{2} + 4y – x – 5

9 + 4y + 4 – x – 1

(y+2)^{2 }– (x-1)

Comparing with (y – k)^{2} – 4a (x – h)

we have h=1, k = -2, a= \(\frac{1}{4}\)

(i) Vertex = (h, k) – (1, -2)

(ii) Focus = (h+a,k) = \(\left(\frac{5}{4},-2\right)\)

(iii) Directrix is x – h + a = 0 ⇒ 4x – 3 = 0

(Iv) Equation of axis is y-k=0 ⇒ y+2 = 0

Question 2.

Find the equation of parabola whose vertex is (3-2) and focus is (3,1)

Solution:

The abscissae of vertex and focus is ‘3’.

Hence the axis of the parabola is x -3, a line parallel to y-axis, focus is above the vertex.

a = SA = \(\sqrt{(3-3)^2+(1+2)^2}=3\)

(∵ A=(3,-2) S= (3,3))

∴ Equation of the parabola

(x-h)^{2} =4a (y-k)

(x-3)^{2}– 12(y+2)

(∵ A = (h,k) =3,-2 and S (h, k-a)]

Question 3.

Find the coordinates of the points on the parabola y^{2} = 2x whose focal distance is \(\frac{5}{2}\)

Solution:

Let P(x_{1}, y_{1}) be any point on the parabola

y^{2} = 2x whose focal distance is \(\frac{5}{2}\) then

\(y_1^2=2 x_1\) and \( x_1+a=\frac{5}{2} \Rightarrow \frac{5}{2}-\frac{1}{2}\)

Where 4a=2 ⇒ a = \(\frac{1}{2}\)

∴ x_{1}= 2, and since y_{12} = 2x_{1} we have

y_{12 }= 4 = y_{1 }= ± 2

∴ The required points are (2, 2), (2, -2).

Short Answer Type Questions

Question 1.

Find the equation of the parabola passing through the points (- 1, 2), (1, – 1) and (2, 1) and having its axis parallel to the x-axis.

Solution:

Since the axis is parallel to x – axis the equation of the parabola is in the form of

x – ly^{2}+my+n ……………. (1)

Given that the parabola passes through the points (-1, 2), (1, -1) and (2, 1) we have

– 1 = 14l+2m+n ……………………. (2)

1= l – m+n ……………….. (3)

2=l+m+n …………………….. (4)

From (2) and (3), -2 = 3l + 3m

∴ Equation of required parabola is

\(x=-\frac{7}{6} y^2+\frac{1}{2} y+\frac{8}{3}\)

Question 2.

A double ordinate of the curve y^{2} = 4ax is of length 8a. Prove that the lines from the vertex to its ends are at right angles.

Solution:

Question 3.

Find the condition for the straight line lx + my + n=0 to be a tangent to the parabola y^{2} = 4ax and find the coordinates

of the point of contact.

Solution:

Let the line ix + my + n = 0 …………….. (1) touches

y^{2} = 4ax …………….. (2) at the point (x_{1}, y_{1}).

Then the equation of tangent at (x_{1}, y_{1}) to the parabola

y^{2}=4ax is yy_{1}– 2ax – 2ax_{1}=0 …………. (2)

Since (1) and (2) represent the same line, coefficients are proportional.

Question 4.

Show that the straight line 7x+6y=13 is a tangent to the parabola y^{2}– 7x – 8y + 14 = 0 and find the point of contact

Solution:

Given line is 7x + 6y – 13 = 0 ……………….. (1)

and parabola is y^{2 }– 7x – 8y+ 14 = 0 ………………..(2)

From (1),

Point of contact is (1, 1) and this point satisfies (1) and (2). Hence (1) is tangent to the parabola (2).

Question 5.

From an external point P, tangents are drawn to the parabola y^{2} = 4ax and these tangents make angles θ_{1}, θ_{2} with its axis such that tan θ_{1}+ tan θ_{2} is a constant b. Then show that P lies on the line y = bx.

Solution:

Let P(x_{1}, y_{1}) be the external point.

Any tangent to the parabola is of the form

y=mx+ \(\frac{\mathrm{a}}{\mathrm{m}}\) =mx_{1}– my_{1}+a = 0

Let the roots of (1) be 4 m_{1}, m_{2} ……………… (1)

Then m_{1} + m_{2} = \(\frac{y_1}{x_1}\) and m_{1} + m_{2 }= \(\frac{\mathrm{a}}{\mathrm{x}_1}\)

The tangents make angles θ_{1} and θ_{2} with the axis such that m_{1} = tan θ_{1} and m_{2}= tan θ_{2}

Given tan θ_{1} + tan θ_{2} = b

⇒ m_{1} + m_{2} = b

\(\frac{\mathrm{y}_1}{\mathrm{x}_1}\) = b ⇒ y_{1 }= bx_{1}

∴ Locus of (x_{1}, y_{2}) is y = bx.

Question 6.

Show that the common tangents to the parabola y^{2} = 4ax and x^{2} = 4by is xa^{1/3 }+ yb^{1/3} + xa^{2/3 }yb^{2/3}

Solution:

The equations of the parabolas are

y^{2 }= 4ax ………………….. (1)

and x^{2 }= 4by ………………… (2)

Equation of any tangent to (1) is of the form

y=mx+\(\frac{\mathrm{a}}{\mathrm{m}}\) ………….. (3)

lf the line (3) is tangent to (2) also the point of intersection of (2) and (3) coincide.

From (3) and (2)

x^{2} = 4b \(\left(\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}\right)\)

⇒ mx^{2} – 4bm^{2}x – 4ab = 0

which should have equal roots.

∴ Therefore the discriminant is zero.

Long Answer Type Questions

Question 1.

Find the equation of a parabola in his standard form.

Solution:

Let S be the focus, l be the directrix as shown in the figure. Let Z be the projection of S on I and A be the mid point of \(\overline{\mathrm{SZ}}\). A lies on the parabola since SA = AZ. A is the vertex of the parabola. YY’ is parallel to the clirectrix ‘l’.

Let S=(a,0), (a>0) then Z =(-a, 0) and the equation of the directrix is x + a = 0.

Let P(x, y) be a point on the parabola and PM is the perpendicular distance from P to x + a = 0.

Then by definition \(\frac{\mathrm{SP}}{\mathrm{PM}}\) = e = 1

= SP^{2} = PM^{2}

= (x- a)^{2 }+(y-0)^{2 }= (x+a)^{2}

= y^{2} =(x+a)^{2}-(x-a)^{2 }= 4ax

Conversely also if P(x, y) is a point such that y^{2} = 4ax then SP = PM.

∴ P(x, y) lies on the locus. Equation of parabola in standard form is y^{2} = 4ax.

Question 2.

(i) If the coordinates of the ends of a focal chord of the parabola y^{2} = 4ax are (x_{1}, y_{1}) and (x_{2}, y_{2}) then prove that

x_{1}x_{2} = a^{2} and y_{1}y_{2} = – 4a^{2}.

(ii) For a focal chord PQ of the parabola y^{2} = 4ax if SP =l and SQ = l’ then prove that \(\frac{1}{l}+\frac{1}{l}=\frac{1}{a}\)

Solution:

(i) Let P(x_{1}, y_{1}) = (at_{1}^{2}, 2at_{1})

and Q(x_{2}, y_{2}) = (at_{2}^{2}, 2at_{2})

be two end points of a focal chord and S(a, 0) be the focus of the parabola y^{2} = 4ax. PSQ is a focal chord P, S, Q are collinear.

(ii) Let P(at^{2}_{1}, 2at_{1}) and Q (at^{2}_{2}, 2at_{2}) be the extremities of a focal chord of parabola such that t_{1},t_{2} = – 1.

Question 3.

If Q is the foot of the perpendicular from a points P on the parabola y^{2} = 8(x -3) to its directrix. S is the focus of parabola and if SPQ is an equilateral triangle then find the length of side of triangle.

Solution:

Given equation of parabola is

(y – 0)^{2} = 8(x – 3)

which is of the form (y – k)^{2} = 4a (x – h)

where 4a = 8 ⇒ a = 2

∴ Vertex = (h, k) = (3,0)

and focus = (h + a, k) = (3 +2, 0) = (5, 0)

Since PQS is an equilateral triangle.

\(\angle \mathrm{SQP}=60^{\circ} \Rightarrow \angle \mathrm{SQZ}=30^{\circ}\)

Also in ΔSZQ , we have sin 30° = \(\frac{\mathrm{SZ}}{\mathrm{SQ}}\)

Hence length of each side of the triangle is 8

Question 4.

The cable of a uniformly loaded suspension bridge hangs In the form of a parabola. The roadway which is horizontal and 72mt. long is supported by vertical wires attached to the cable. The longest being 30 mts. and the shortest being 6 mts. Find the length of the supporting wire attached to the roadway 18 mts. from the middle.

Solution:

Let AOB be the cable [O is the lowest point and A, B are highest points). Let PRQ be the suspension bridge suspended with PR – RQ = 36 mts.

PA = QB = 30 mts (longest wire)

OR = 6 mts (shortest wire)

∴ PR = RQ = 36 mts. We take O as origin of coordinates at O, X-axis along the tangent \(\overline{\mathrm{RO}}\). So the equation of the cash is x^{2} = 4ay for some a> 0.

∴ B(36,24) and B is a point on x^{2}=4ay

We have (36)^{2} = 4a(24)

⇒ 4a = \(\frac{36 \times 36}{24}\) = 54 mts

If RS = 18 mts and SC is the vertical through S meeting the cable at ‘C’ and the X- axis at D.

Then SC is the length of the required wire.

Let SC = l mts then DC = L – 6 mts.

∴ C = (18, 1 – 6) which lies on x^{2} = 4ay

⇒ (18)^{2} = 4a (l – 6)

⇒ (18)^{2}= 54 (l -6)

⇒ l – 6 = \(\frac{18 \times 18}{54}\) = 6

⇒ l = 12.

Question 5.

Prove that the normal chord at the point other than origin whose ordinate is equal to Its abscissa subtends a right angle at the focus

Solution:

Let the equation of the parabola

y^{2 }= 4ax ……………. (1)

and P(at^{2}, 2at) be any point

Ordinate – abscissa

= 2at = at^{2} = t=0 or t = 2

But t ≠ 0, Hence the point (4a, 4a) at which normal is

⇒ y + 2x = 2a(2) + a(2)^{3}

⇒ y + 2x = 4a + 8a

⇒ y=12a -2x …………….. (2)

∴ From (1) we get

(12a – 2x)^{2 }– 4ax

⇒ 4x^{2} – 48ax + 144a^{2} = 4ax

⇒ 4x^{2} – 52ax+ 144a^{2}= 0

⇒ x^{2} – 13ax + 36a^{2} = 0

⇒ (x – 4a) (x – 9a) – 0

⇒ x = 4a, 9a

Correspondingly we get y = 12a – 8a = 4a and y = 12a – 18a= – 6a

Hence the other points of intersection of normal at P(4a, 4a) is Q(9a – 6a).

We have S = (a, 0)

Question 6.

Prove that the area of the triangle formed by the tangents at (x_{1}, y_{1}), (x_{2}, y_{2}) and (x_{3},y_{3}) to the parabola y^{2 }= 4ax

\(\frac{1}{16 a}\left|\left(y_1-y_2\right)\left(y_2-y_3\right)\left(y_3-y_1\right)\right|\) sq. unit

Solution:

Let A (x_{1}, y_{1}) = (at_{2}^{2}, 2at_{1})

B (x_{2}, y_{2}) = (at_{2}^{2}, 2at_{2})

C (x_{3}, y_{3}) – (at_{3}^{2}, 2at_{3}) be the three points on the parabola y^{2}=4ax.

The equations of tangents at A, B, C are

∴ Point of intersection of tangents at A and B is P(at_{1}t_{2}, a(t_{1} + t_{2})).

Similarly the points of intersection of tangents at B and C as well as A & C are

Question 7.

Prove that the two parabolas y^{2} = 4ax and x^{2} = 4by intersect other than the origin) at tan^{-1}

\(\left(\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right)\)

Solution:

Take a>0, b>0

and y^{2}=4ax ………………….. (1)

and x^{2} = 4by are ……………. (2)

the given parabolas.

Solving (1) and (2) we get the point of intersection other than the origin.

Question 8.

Prove that the orthocentre of the triangle formed by any three tangents to a parabola lies on the directrix of the parabola.

Solution:

Let y^{2} = 4ax be the parabota and

A = (at_{1}^{2}, 2at_{1}), B = (at_{2}^{2}, 2at_{2}),

C = (at_{3}^{2}, 2at_{3}) be any three points on It.

If P, Q, R are the points of intersection of tangents at A and B, B and C, C and A then

P = [at_{1}t_{2}, a(t_{1} + t_{2})] Q = [at_{2}t_{3}, a(t_{2} + t_{3})] R = [at_{1}t_{3}, a(t_{1} + t_{3})]

Consider the ΔPQR then equation \(\overline{\mathrm{QR}}\) of (Tangent at C) is

x – yt_{3} + at_{3}^{2} = 0

∴ Altitude through P of Δ PQR is

t_{3}x + y = at_{1}t_{2}t_{3} + a(t_{1} + t_{2}) …………….. (1)

∴ Slope = \(\frac{1}{t_3}\) and equation is

y – a(t_{1}+ t_{2}) = – t_{3} [x – at_{1}t_{2}]

y+xt_{3 }– at_{1}t_{2}t_{3}+a(t_{1} +t_{2})

Similarly the altitude through Q is

t_{1}x + y – at_{1}t_{2}t_{3} + a(t_{2}+ t) …………… (2)

Solving (1) and (2)

x(t_{3 }– t_{1}) = a(t_{1}– t)

⇒ x = – a

Hence the orthocentre of ΔPQR with x coordinate as – ‘a’ lies on the directrix of the parabola.