Students get through Maths 2B Important Questions Inter 2nd Year Maths 2B Parabola Important Questions which are most likely to be asked in the exam.

## Intermediate 2nd Year Maths 2B Parabola Important Questions

Question 1.

Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2). [T.S. Mar. 15]

Solution:

Let S = (1, -7), A(1, -2)

h = 1, k = -2, a = -2 + 7 = 5

The Axis of the parabola is parallel to the y-axis

The equation of the parabola is

(x – h)^{2} = – 4a (y – k)

(x – 1)^{2} = – 20(y + 2)

x^{2} – 2x + 1, = – 20y – 40

⇒ x^{2} – 2x + 20y + 41 – 0.

Question 2.

If the normal at the point t_{1} on the parabola y^{2} =, 4ax meets it again at point t^{2} then prove that t_{1}t_{2} + t_{1}^{2} + 2 = 0. [May 07]

Solution:

Equation of normal is

y – y_{1} = \(\frac{-y_{1}}{2 a}\) (x – x_{1})

y – 2at_{1} = \(\frac{-2 a t_{1}}{2 a}\) (x – at_{1}^{2}) ……………… (i)

Equation of the line (i) again meets parabola at (at_{2}^{2}, 2at_{1})

∴ 2at_{2} – 2at_{1} = t_{1} (at_{2}^{2} – at_{1}^{2})

\(-\frac{2}{t_{1}}\) = t_{1} + t_{2} ⇒ -2 = t_{1}^{2} + t_{1}t_{2}

⇒ t_{1}^{2} + t_{1}t_{2} + 2 = 0

Question 3.

Find the coordinates of the vertex and focus, and the equations of the directrix and axes of the y – x + 4y + 5 = 0. [Mar 05]

Solution:

y^{2} – x + 4y + 5 = 0 ⇒ (y – (-2))^{2} = (x – 1), comparing with (y – k)^{2} = 4a(x – h),we get (h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)

coordinates of the focus (h + a, k) = (\(\frac{5}{4}\), -2)

Equation of the directrix x – h + a = 0

i.e., 4x – 3 = 0

Equation of the axis y – k = 0.

i.e., y + 2 = 0

Question 4.

Find the coordinates of the points on the parabola y^{2} = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]

Solution:

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

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

y_{1}^{2} = 2x_{1} and x_{1} + a = \(\frac{5}{2}\)

⇒ x_{1} + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x_{1} = 2

∴ y_{1}^{2} = 2(2) = 4 ⇒ y_{1} = ±2

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

Question 5.

Find the value of k if the line 2y = 5x + k is a tangent to the parabola y^{2} = 6x. [T.S. Mar. 16]

Solution:

Given line is 2y = 5x + k

⇒ y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\))

Comparing y = (\(\frac{5}{2}\))x + (\(\frac{k}{2}\)) with y = mx + c

We get m = \(\frac{5}{2}\), c = \(\frac{k}{2}\)

y = (\(\frac{5}{2}\))x + \(\frac{k}{2}\) is a tangent to y^{2} = 6x

c = \(\frac{a}{m}\)

⇒ \(\frac{k}{2}=\frac{\left(\frac{3}{2}\right)}{\left(\frac{5}{2}\right)}\) ⇒ k = \(\frac{6}{5}\)

Question 6.

Show that the equations of common tangents to the circle x^{2} + y^{2} = 2a^{2} and the parabola y^{2} = 8ax are y = ± (x + 2a). [Mar. 06]

Solution:

The equation of tangent to parabola

y^{2} = 8ax is y = mx + \(\frac{2a}{m}\)

m^{2}x – my + 2a = 0 ……………….. (1)

If (i) touches circle x^{2} + y^{2} = 2a^{2}, then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a\(\sqrt{2}\) of the circle.

\(\left|\frac{2 a}{\sqrt{m^{2}+m^{4}}}\right|=a \sqrt{2}\)

or 4 = 2 (m^{4} + m^{2})

m^{4} + m^{2} – 2 = 0

(m^{2} + 2) (m^{2} – 1) = 0 or m = ±1

Required tangents are

y = (1) x + \(\frac{2 a}{(1)}\), y = (-1) x + \(\frac{2 a}{(-1)}\)

y = ± (x + 2a)

Question 7.

Find the co-ordinates of the point on the parabola y^{2} = 8x whose focal distance is 10. [T.S. Mar. 17] [A.P. Mar. 16; Mar. 14, 11]

Solution:

Equation of the parabola is

y^{2} = 8x

4a = 8 ⇒ a – 2

Co-ordinates of the focus S are (2, 0) Suppose P(x, y) is the point on the parabola.

Given SP = 10 ⇒ SP^{2} = 100

(x – 2)^{2} + y^{2} = 100

But y^{2} = 8x

⇒ (x – 2)^{2} + 8x = 100

⇒ x^{2} – 4x + 4 + 8x – 100 = 0

⇒ x^{2} + 4x – 96 = 0 ⇒ (x + 12) (x – 8) = 0

x + 12 = 0 or x – 8 = 0

x = -12, or 8

Case (i) x = 8

y^{2} = 8.x = 8.8 = 64

y = ±8

Co-ordinates of the required points are (8, 8) and (8, -8)

Case (ii) x = -12

y^{2} = 8(-12) = -96 < 0

y is not real.

Question 8.

If (\(\frac{1}{2}\), 2) is one extermity of a focal chord of the parabola y^{2} = 8x. Find the co-ordinates of the other extremity. [May 06]

Solution:

A = (\(\frac{1}{2}\), 2); S = (2, 0)

B = (x_{1}, y_{1}) ⇒ (\(\frac{y_{1}^{2}}{8}\), y_{1})

ASB is a focal chord.

∴ Slopes of SA and BS are same.

or 4y_{1}^{2} + 24y_{1} – 64 = 0

⇒ y_{1}^{2} + 6y_{1} – 16 = 0

⇒ (y_{1} + 8) (y, – 2) = 0

y_{1} = 2, 8

x_{1} = \(\frac{1}{2}\), 8; So (8, -8) other extremity.

Question 9.

Show that the common tangent to the parabola y^{2} = 4ax and x^{2} = 4by is xa^{1/3} + yb^{1/3} + a^{2/3}b^{2/3} = 0 [T.S. Mar. 17] [A.P. Mar. 16]

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 + \(\frac{a}{m}\) …………….. (3)

If the line (3) is a tangent to (2) also, we must get only one point of intersection of (2) and (3).

Substituting the value of y from (3) in (2), we get x^{2} = 4b (mx + \(\frac{a}{m}\)) is 3x^{2} – 4bm^{2}x – 4ab = 0 should have equal roots Therefore its discriminent must be zero. Hence

16b^{2}m^{4} – 4m (-4ab) = 0

16b(bm^{4} + am) = 0

m(bm^{3} + a) = 0 But m ≠ 0

∴ m = – a^{1/3}/b^{1/3} substituting in (3) the equation of the common tangent becomes

y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) or

a^{1/3}x + b^{1/3}y + a^{2/3}b^{2/3} = 0 .

Question 10.

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 (a > 0) is \(\frac{1}{16a}\)|(y_{1} – y_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})| sq. units. [T.S. Mar. 15]

Solution:

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

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

and F(x_{3}, y_{3}) = (at_{3}^{2}, 2at_{3}), be three points on the parabola

y^{2} = 4ax (a > 0).

The equation of the tangents at D, E and F are

t_{1}y = x + at_{1}^{2} ……………… (1)

t_{2}y = x + at_{2}^{2} ………………. (2)

t_{3}y = x.+ at_{3}^{2} ………………. (3)

(1) – (2) ⇒ (t_{1} – t_{2}) y = a(t_{1} – t_{2}) (t_{1} + t_{2})

⇒ y = a(t_{1} + t_{2}) substituting in (1)

we get x = at_{1}t_{2}.

∴ The point of intersection of the tangents at D and E is say P(at_{1}t_{2}, a(t_{1} + t_{2}))

Similarly the points of intersection of tangent at E, F and at F, D are Q(at_{2}t_{3}, a(t_{2} + t_{3}) and R(at_{3}t_{1} a(t_{3} + t_{1})) respectively

Area of ∆PQR

\(\frac{1}{16a}\) |2a(t_{1} – t_{2}) 2a(t_{2} – t_{3}) 2a(t_{3} – t_{1})|

\(\frac{1}{16a}\)|(y_{1} – y_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})| sq. units.

Question 11.

Prove that the two parabolas y^{2} = 4ax and x^{2} = 4by intersect (other than the origin) at an angle of Tan^{-1} \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]

Solution:

Without loss of generality we assume a > 0 and b > 0.

Let P(x, y) be the point of intersection of the parabolas other than the origin. Then

y^{4} = 16a^{2}x^{2}

= 16a^{2}(4by)

= 64a^{2}by

∴ y[y^{3} – 64a^{2}b] = 0

=> y^{3} – 64a^{2}b = 0

=> y = (64a^{2}b)^{1/3} [∵ y > 0]

= 4a^{2/3}b^{1/3}

Also from y^{2} = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)

= 4a^{1/3}b^{2/3}

∴ P = (4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})

Differentiating both sides of y^{2} 4ax w.r.t ‘x’, we get

\(\frac{d y}{d x}=\frac{2 a}{y}\)

∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)

If m_{1} be the slope of the tangent at P to y^{2} = 4ax, then

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

Similarly, we get m_{2} = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m_{2} is the slope of the tangent at P to x^{2} = 4by.

If θ is the acute angle between the tangents to the curves at P, then

tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)

so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 12.

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

(i) y^{2} = 16x

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

(iii) 3x^{2} – 9x + 5y – 2 = 0

(iv) y^{2} – x + 4y + 5 = 0 [Mar. 05]

Solution:

i) y^{2} = 16x, comparing with y^{2} = 4ax,

we get 4a = 16 ⇒ a = 4

The coordinates of the vertex = (0, 0)

The coordinates of the focus = (a, 0) = (4, 0)

Equation of the directrix: x + a = i.e., x + 4 = 0

Axis of the parabola y = 0

ii) x^{2} = -4y, comparing with x^{2} = -4ay,

we get 4a = 4 ⇒ a = 1

The coordinates of the vertex = (0, 0

The coordinates of the focus = (0, -a) = (0, 1)

The equation of the directrix y – a = 0

i.e., y – 1 = 0 .

Equation of the axis x = 0

iii) 3x^{2} – 9x + 5y – 2 = 0

3(x^{2} – 3x) = 2 – 5y

⇒ 3(x^{2} – 2x(\(\frac{3}{2}\)) + \(\frac{9}{4}\)) = 2 – 5y + \(\frac{27}{4}\)

(x – \(\frac{3}{2}\))^{2} = –\(\frac{5}{3}\) (y – \(\frac{7}{4}\)),

Comparing with (x – h)^{2} = -4a (y – k) we get

a = \(\frac{5}{12}\), h = \(\frac{3}{2}\), k = \(\frac{7}{4}\)

∴ Coordinates of the vertex = (h, k)

= (\(\frac{3}{2}\), \(\frac{7}{4 }\))

Coordinates of the focus = (h, k – a)

= (\(\frac{3}{2}\), \(\frac{7}{4}\) – \(\frac{5}{12}\)) = (\(\frac{3}{2}\), \(\frac{4}{3}\))

Equation of the directrix is y – k – a = 0

i.e., 6y – 13 = 0

Equation of the axis is x – h = 0

i.e., 2x – 3 = 0

iv) y^{2} – x + 4y + 5 = 0 ⇒ (y- (-2))^{2} = (x – 1),

comparing with (y – k)^{2} = 4a(x – h),we get

(h, k) = (1, -2) and a = \(\frac{1}{4}\), coordinates of the vertex (h, k) = (1 ,-2)

coordinates of the focus

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

Equation of the directrix x – h + a = 0

i.e., 4x – 3 = 0

Equation of the axis y – k = 0.

i.e., y + 2 = 0

Question 13.

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

Solution:

The abcissae of the vertex and focus are equal to 3. Hence the axis of the parabola is x = 3, a line parallel to y-axis, focus is above the vertex.

a = distance between focus and vertex = 3.

∴ Equation of the parabola

(x – 3)^{2} = 4(3) (y + 2)

i.e., (x – 3)^{2} = 12(y + 2).

Question 14.

Find the coordinates of the points on the parabola y^{2} = 2x whose focal distance is \(\frac{5}{2}\). [A.P. Mar. 15, Mar. 13]

Solution:

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

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

y_{1}^{2} = 2x_{1} and x_{1} + a = \(\frac{5}{2}\)

⇒ x_{1} + \(\frac{1}{2}\) = \(\frac{5}{2}\) ⇒ x_{1} = 2

∴ y_{1}^{2} = 2(2) = 4 ⇒ y_{1} = ±2

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

Question 15.

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. .

Since the parabola passes through (-1, 2), we have

-1 = l( 2)^{2} + m(2) + n

⇒ 4l + 2m + n = – 1 ……………………. (1)

Similarly, since the parabola passes through (1,-1) and (2, 1) we have

l – m + n = 1 ……………… (2)

l + m + n = 2 ……………… (3)

Solving (1), (2) and (3)

we get l = –\(\frac{7}{6}\), m = \(\frac{1}{2}\) and n = \(\frac{8}{3}\).

Hence the equation of the parabola is

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

7y^{2} – 3y + 6x- 16 = 0.

Question 16.

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

Solution:

Let P = (at^{2}, 2at) and P’ = (at^{2}, -2at) be the ends of double ordinate PP’. Then

8a = PP’ = \(\sqrt{0+(4 a t)^{2}}\) = 4at ⇒ t = 2.

∴ P = (4a, 4a), P’= (4a, -4a) .

Slope of \(\overline{\mathrm{AP}}\) × slope of \(\overline{\mathrm{AP}^{\prime}}\)

= (\(\frac{4a}{4a}\))(-\(\frac{4a}{4a}\)) = -1

∴ ∠PAP’ = \(\frac{\pi}{2}\)

Question 17.

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}, 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_{1}) be two end points of a focal chord.

P, S, Q are collinear

Slope of \(\overline{\mathrm{PS}}\) = Slope of \(\overline{\mathrm{QS}}\)

\(\frac{2 a t_{1}}{a t_{1}^{2}-a}=\frac{2 a t_{2}}{a t_{2}^{2}-a}\)

t_{1}t_{2}^{2} – t_{1} = t_{2}t_{1}^{2} – t_{2}

t_{1}t_{2} (t_{2} – t_{1}) + (t_{2} – t_{1}) = 0

1 + t_{1}t_{2} = 0 ⇒ t_{1}t_{2} = -1 ………………… (1)

From (1) x_{1}x_{2} = at_{1}^{2} at_{2}^{2} = a^{2}(t_{2}t_{1})^{2} = a^{2}

y_{1}y_{2} = 2at_{1}2at_{2} = 4a^{2}(t_{2} t_{1}) = -4a^{2}

ii) Let P(at_{1}^{2}, 2at_{1}) and Q(at_{2}^{2}, 2at_{1}) be the extremities of a focal chord of the parabola, then t_{1}t_{2} = -1 (from (1))

Question 18.

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

Solution:

Given parabola y^{2} = 8(x – 3) then

its vertex A = (3, 0) and focus = (5, 0)

[4a = 8 ⇒ a = 2] since PQS is an equilateral triangle

Hence length of each side of triangle is ‘8’.

Question 19.

The cable of a uniformely loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and 72 mt. 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 road-way 18 mts. from the middle.

Solution:

Let AOB be the cable [0 is its lowest point and A, B are the highest points]. Let PRQ be the bridge suspended with PR = RQ = 36 mts. (see above Fig.).

PA = QB = 30 mts (longest vertical sup-porting wires)

OR = 6 mts (shortest vertical supporting wire) [the lowest point of the cable is up¬right the mid-point R of the bridge]

Therefore, PR = RQ = 36 mts. We take the origin of coordinates at 0, X-axis along the tangent at O to the cable and the Y-axis along \(\overleftrightarrow{\mathrm{RO}}\). The equation of the cable would, therefore, be x^{2} = 4ay for some a > 0. We get B = (36, 24) and 36^{2} = 4a × 24.

Therefore, 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 supporting wire required. If SC = l mts, then

DC = (l – 6) mts. As such C = (18, l – 6).

Since C is on the cable, 18^{2} = 4a (l – 6)

⇒ l – 6 = \(\frac{18^{2}}{4 a}\) = \(\frac{18 \times 18}{54}\) = 6

⇒ l = 12

Question 20.

Find the condition for the straight line lx + my + n = 0 to be a tangent to the parabola y^{2} = 4ax and find the co-ordinates of the point of contact.

Solution:

Let the line lx + my + n = 0 be a tangent to the parabola y^{2} = 4ax at (at^{2}, 2at). Then the equation of the tangent at P(t) is x – yt + at^{2} = 0 then it represents the given line

lx + my + n = 0, then

Question 21.

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:

Equation of the given line is 7x + 6y = 13, equation of the given parabola is y^{2} – 7x – 8y + 14 = 0.

By eliminating x, we get the ordinates of the points of intersection of line and parabola adding the equations y^{2} – 2y + 1 = 0.

i.e., (y – 1 )^{2} = 0 ⇒ y = 1, 1.

∴ The given line is tangent to the given parabola.

If y = 1 then x = 1 hence the point of contact is (1, 1).

Question 22.

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 be

y^{2} = 4ax and P(at^{2}, 2at) be any point …………………. (1)

On the parabola for which the abscissa is equal to the ordinate.

i.e., at^{2} = 2at ⇒ t = 0

or t = 2. But t ≠ 0. Hence the point (4a, 4a) at which the normal is

y + 2x = 2a(2) + a(2)^{3} (or)

y = (12a – 2x) ………………….. (2)

Substituting the value of

y = 12a – 2x in (1) we get

(12a – 2x)^{2} = 4ax (or)

x^{2} – 13ax + 36a^{2} = (x – 4a) (x – 9a) = 0

⇒ x = 4a, 9a

corresponding values of y are 4a and -6a. Hence the other points of intersection of that normal at P(4a, 4a) to the given parabola is Q(9a, -6a), we have S(a, 0).

Slope of the \(\overline{\mathrm{SP}}\) = m_{1} = \(\frac{4a-0}{4a-a}\) = \(\frac{4}{3}\),

Slope of the \(\overline{\mathrm{SQ}}\) = m_{2} = \(\frac{-6a-0}{9a-a}\) = \(\frac{3}{4}\)

clearly m_{1}m_{2} = -1, so that \(\overline{\mathrm{SP}}\) ⊥ \(\overline{\mathrm{SQ}}\)

Question 23.

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

Solution:

Let the coordinates of P be (x_{1}, y_{1}) and the equation of the parabola y^{2} = 4ax. Any tangent to .the parabola is y = mx + \(\frac{a}{m}\), if this passes through (x_{1}, y_{1}) then

y_{1} = mx, + \(\frac{a}{m}\)

i.e., m^{2}x_{1} – my_{1} + a = 0 ………………… (1)

Let the roots of (1) be m_{1}, m_{21}.

Then m_{1} + m_{2} = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)

⇒ tan θ_{1} + tan θ_{2} = \(\frac{\mathrm{y}_{1}}{\mathrm{x}_{1}}\)

[∵ The tangents make angles θ_{1}, θ_{2} with its axis (x – axis) then their slopes m_{1} = tan θ_{1} and m_{2} = tan θ_{2}].

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

∴ P(x_{1}, y_{1}) lies on the line y = bx.

Question 24.

Show that the common tangent to the parabola y^{2} = 4ax and x^{2} = 4by is xa^{1/3} + yb^{1/3} + a^{2/3}b^{2/3} = 0. [A.P. Mar. 16]

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 + \(\frac{a}{m}\)

If the line (3) is a tangent to'(2) also, we must get only one point of intersection of (2) and (3).

Substituting the value of y from (3) in (2),

we get x^{2} = 4b (mx + \(\frac{a}{m}\)) is mx^{2} – 4bm^{2}x – 4ab = 0 should have equal roots therefore its discriminent must be zero. Hence

16b^{2}m^{4} – 4m (-4ab) = 0

16b(bm^{4} + am) = 0

m(bm^{3} + a) = 0 But m ≠ 0

∴ m = – a^{1/3}/b^{1/3} substituting in (3) the equation of the common tangent becomes

y = \(-\left(\frac{a}{b}\right)^{1 / 3} x+\frac{a}{\left(-\frac{a}{b}\right)^{1 / 3}}\) (or)

∴ a^{1/3} x + b^{1/3}y + a^{2/3}b^{2/3} = 0

Question 25.

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 (a > 0) is \(\frac{1}{16a}\)|(y_{1} – y_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})| sq. units. [T.S. Mar. 15]

Solution:

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

E(x_{2}, y_{2}) = (at_{3}^{2}, 2at_{2})

and F(x_{3}, y_{3}) = (at_{2}^{3}, 2at_{3})

be three points on the parabola

y^{2} = 4ax (a > 0).

The equation of the tangents at D, E and F are

t_{1}y = x + at_{1}^{2} ………………. (1)

t_{2}y = x + at_{2}^{2} ………………. (2)

t_{3}y = x + at_{3}^{2} ………………. (3)

(1) – (2) ⇒ (t_{1} – t_{2}) y = a(t_{1} – t_{2}) (t_{1} + t_{2})

⇒ y = a(t_{1} + t_{2}) substituting in (1)

we get x = at_{1}t_{2}

∴ The point of intersection of the tangents at D and E is say P(at_{1}t_{2}, a(t_{1} + t_{2}))

Similarly the points of intersection of tangent at E, F and at F, D are Q(at_{2}t_{3}, a(t_{2} + t_{3}) and R(at_{3}t_{1}, a(t_{3} + t_{1})) respectively

Area of ∆PQR

Question 26.

Prove that the two parabolas y^{2} = 4ax and x^{2} = 4by intersect (other than the origin) at an angle of Tan^{-1} \(\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\) [Mar. 14]

Solution:

Without loss of generality we assume a > 0 and b > 0.

Let P(x, y) be the point of intersection of the parabolas other than the origin. Then

y^{4} = 16a^{2}x^{2}

= 16a^{2}(4by)

= 64a^{2}by

∴ y[y^{3} – 64a^{2}b] = 0

⇒ y^{3} – 64a^{2}b = 0

⇒ y = (64a^{2}b)^{1/3} [∵ y > 0]

= 4a^{2/3}b^{1/3}

Also from y^{2} = 4ax, x = \(\frac{16 a^{4 / 3} b^{2 / 3}}{4 a}\)

= 4a^{1/3}b^{2/3}

∴ P = (4a^{1/3}b^{2/3}, 4a^{2/3}b^{1/3})

Differentiating both sides of y^{2} 4ax w.r.t ‘x’, we get

\(\frac{d y}{d x}=\frac{2 a}{y}\)

∴ \(\left[\frac{\mathrm{dy}}{\cdot \mathrm{dx}}\right]_{\mathrm{p}}=\frac{2 \mathrm{a}}{4 \mathrm{a}^{2 / 3} \mathrm{~b}^{1 / 3}}=\frac{1}{2}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{1 / 3}\)

If m_{1} be the slope of the tangent at P to y^{2} = 4ax, then

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

Similarly, we get m_{2} = \(2\left(\frac{a}{b}\right)^{1 / 3}\) where m_{2} is the slope of the tangent at P to x^{2} = 4by.

If θ is the acute angle between the tangents to the curves at P, then

tan θ = \(\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|=\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\)

so that θ = \(\tan ^{-1}\left[\frac{3 a^{1 / 3} b^{1 / 3}}{2\left(a^{2 / 3}+b^{2 / 3}\right)}\right]\)

Question 27.

Prove that the orthocenter 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 parabola 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.

Now we consider the triangle PQR formed by the tangents to the parabola at A, B, C

where P = (at_{1}t_{2}, a(t_{1} + t_{2})),

Q = (at_{2}t_{3}, a(t_{2} + t_{3})) and R = (at_{3}t_{1}, a(t_{3} + t_{1})).

Equation of \(\overleftrightarrow{\mathrm{QR}}\) (i.e., the tangent at C) is x – t_{3} y + at_{3}^{2} = 0.

Therefore, the attitude through P of triangle PQR is

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

Similarly, the attitude through Q is

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

Solving (1) and (2), we get (t_{3} – t_{1})

x’ = a(t_{1} – t_{3}) i.e., x = – a.

Therefore, the orthocenter of the triangle PQR, with abscissa as -a, lies on the directrix of the parabola.