Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(b) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Parabola Solutions Exercise 3(b)

I.

Question 1.

Find the equation of the tangent and normal to the parabola y² = 6x at the positive end of the latus rectum.

Solution:

(a, 2a) Here 4a = 6 ⇒ a = \(\frac{3}{2}\)

(\(\frac{3}{2}\), 3)

equation of tangent yy_{1} = 2a (x + x_{1})

yy_{1} = 3(x + x_{1})

3y = 3(x + \(\frac{3}{2}\))

2y – 2x – 3 = O’ is the equation of tangent

Slope of tangent is 1

Slope of normal is -1

Equation of normal is y – 3 = -1(x – \(\frac{3}{2}\))

2x + 2y – 9 = 0

Question 2.

Find the equation of the tangent and normal to the parabola x² – 4x – 8y + 12 = 0 at (4, \(\frac{3}{2}\))

Solution:

(x – 2)² – 4 – 8y + 12 = 0

⇒ (x – 2)² – 8y + 8 = 0

⇒ (x – 2)² = 8(y – 1); 4a = 8 ⇒ a = 2

Equation of tangents at (x_{1}, y_{1}) is

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

⇒ (x – 2) (4 – 2) = 2(2) (y – 1 + \(\frac{3}{2}\) – 1)

⇒ 2(x – 2) = 4(\(\frac{2y-1}{2}\)

x – 2y – 1 = 0

Equation of normal will be

y – y_{1} = m(x – x_{1})

m – Slope of normal

Slope of tangent is \(\frac{1}{2}\)

Slope of normal is – 2

y – \(\frac{3}{2}\) = -2(x – 4)

2y – 3 = -4x + 16

4x + 2y – 19 = 0

Question 3.

Find the value of k if the line 2y = 5x + k is a tangent to the parabola y² = 6x.

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 y² = 6x

Question 4.

Find the equation of the normal to the parabola y² = 4x which is parellel to y – 2x + 5 = 0.

Solution:

Equation of the parabola is y² = 4x

∴ a = 1

The equation of the given line y – 2x + 5 = 0

Slope m = 2

The normal is parallel to the line y – 2x +5 = 0

Slope of the normal = 2

Equation of the normal at ‘t’ is

y + tx = 2at + at³

∴ Slope = -t = 2

⇒ t = -2

Equation of the normal is

y – 2x = 2.1 (-2) + 1(-2)³

= -4 – 8 = – 12

2x-y- 12 = 0.

Question 5.

Show that the line 2x – y + 2 = 0isa tangent to the parabola y² = 16x. Find the point of contact also. .

Solution:

Given line is 2x – y + 2 = 0

⇒ y = 2x + 2

Comparing with y = mx + c we get m = 2, c = 2

Comparing y² = 16x with y² = 4ax

We get 4a = 16 ⇒ a = 4

\(\frac{a}{m}=\frac{4}{2}\) = 2 = c

Point of contact = (\(\frac{a}{m^{2}},\frac{2a}{m}\)) = (\(\frac{4}{2^{2}},\frac{2(4)}{2}\))

= (1.4)

Question 6.

Find the equation of tangent to the parabola y2 = 16x inclined at an angle 60° with its axis and also find the point of contact.

Solution:

θ = 60°; m = tan 60° = √3

II.

Question 1.

Find the equations of tangents to the parabola y² = 16x which are parallel and perpendicular respectively to the line 2x – y + 5 = 0, also find the co-ordinates of the points of contact also.

Solution:

Equation of the parabola is y² = 16x

The tangent is parallel to 2x – y + 5 = 0

Equation of the tangent can be taken as y = 2x + c

Condition for tangency is c = \(\frac{a}{m}=\frac{4}{2}\) = 2

Equation of the parallel tangent is y = 2x + 2

2x – y + 2 = 0

Point of contact is (\(\frac{a}{m^{2}},\frac{2a}{m}\)) = (\(\frac{4}{4},\frac{8}{2}\)) = (1, 4)

Slope of the perpendicular tangept is

m’ = –\(\frac{1}{m}=-\frac{1}{2}\)

Equation of the perpendicular tangent is

y = m’x + c’

= (-\(\frac{1}{2}\)) x + c’

where c’ = \(\frac{a}{m’}\) = \(\frac{4}{\left(-\frac{1}{2}\right)}\) = -8

Equation of the perpendicular tangent is

y = –\(\frac{1}{2}\)x – 8

2y = – x – 16

x + 2y + 16 = 0

Question 2.

If lx + my + n = 0 is a normal to the parabola y² = 4ax, then show that al³ + 2alm² + nm² = 0.

Solution:

Equation of the parabola is y² = 4ax

Equation of the normal is y + tx = 2at + at³

tx + y – (2at + at³) = 0 ……….. (1)

Equation of the given line is

lx + my + n = 0 …………. (2)

(1), (2) represent the same line.

Comparing the co-efficients

Multiplying with m³, we get

-nm² = 2alm² + al³

⇒ al³ + 2alm² + nm² = 0

Question 3.

Show that the equations of common tangents to the circle x² + y² = 2a² and the parabola y² = 8ax are y = ± (x + 2a).

Solution:

The equation of tangent to parabola

y² = 8ax is y = mx + \(\frac{2a}{m}\) ……….. (i)

m²x – my + 2a = 0

If (i) touches circle x² + y² = 2a², then the length of perpendicular from its centre (0, 0) to (i) must be equal to the radius a√2 of the circle.

\(\frac{2a}{\sqrt{m^{2} + m^{4}}}\) = a√2

or 4 = 2 (m^{4} + m²)

m^{4} + m² – 2 = 0

(m² + 2) (m² – 1) = 0 or m = ±1

Required tangents are

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

⇒ y = ±(x + 2a)

Question 4.

Prove that the tangents at the extremities of a focal chord of a parabola intersect at right angles on the directrix.

Solution:

Equation of the parabola is y² = 4ax

Equation of the tangent at Q(_{1}) is

t_{1}y = x + at²_{1}

Equation of the tangent at R(t_{2}) is t_{2}y = x + at²_{2}

Solving, point of intersection is T [at_{1}t_{2}, a(t_{1} + t_{2})]

Equation of the chord QR is

(t_{1} + t_{2})y = 2x + 2at_{1}t_{2}

Question 5.

Find the condition for the line y = mx + c to be a tangent to the parabola x² = 4ay.

Solution:

Equation of tangent to x² = 4ay in terms of ‘m_{1}‘ [slope of tangent] is

Comparing (i) and (ij), we get

c = – am² is required condition.

Question 6.

Three normals are drawn (k, 0) to the parabola y² = 8x one of the normal is the axis and the axis and the remaining two normals are perpendicular to each other, then find the value of k.

Solution:

Equation of any normal to the parabola is

y + xt = 2at + at³

This normal passes through (k, 0)

∴ kt = 2at + at³

at³ + (2a – k)t = 0

at² + (2a – k) = 0

Given m_{1} = 0, m_{2}m_{3} = -1

(-t_{2}) (-t_{3}) = -1 ⇒ t_{2}t_{3}

\(\frac{2a-k}{a}\) = -1

2a – k = -a

k = 2a + a = 3a

Equation of the parabola is

y² = 8x

4a = 8

⇒ a = 2

k = 3a = 3(2) = 6.

Question 7.

Show that the locus of point of intersection of perpendicular tangents to the parabola y² = 4ax is the directrix x + a = 0.

Solution:

Equation of any tangent to the parabola can be taken as y = mx + \(\frac{a}{m}\)

This tangent passes through P(x_{1}, y_{1})

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

my_{1} = m²x_{1} + a

m²x_{1} – my_{1} + a = 0

The tangents are perpendicular

⇒ m_{1}m_{2} = -1

\(\frac{a}{x_1}\) = -1

x_{1} = -a

Locus of P(x_{1}, y_{1}) is x = -a, the directrix of the parabola.

Question 8.

Two Parabolas have the same vertex and equal length of latus rectum such that their axes are at right angle. Prove that the common tangents touch each at the end of latus rectum.

Solution:

Equations of the parabolas can be taken as

y² = 4ax

and x² = 4ay

Equation of the tangent at (2at, at²) to

x² = 4ay is

2atx = 2a(y + at²)

y = tx – at²

This is a tangent to y² = 4ax

∴ The condition is c = \(\frac{a}{m}\)

-at² = \(\frac{a}{t}\)

t³ = -1 ⇒ t = -1

Equation of the tangent is y = -x – a

x + y + a = 0

Equation of the tangent at L’ (a, – 2a) is

y(-2a) = 2a(x + a)

x + y + a = 0

∴ Common tangent to the parabolas touches the parabola y² = 4ax at L (a, -2a) Equation of the tangent at L (-2a, a) to

x² = 4ay

x(-2a) = 2a(y + a)

x + y + a = 0

Common tangent to the parabolas touch the parabola at L’ (-2a, a)

Question 9.

Show that the foot of the perpendicular from focus to the tangent of the parabola y² = 4ax lies on the tangent at vertex.

Solution:

Equation of any tangent to the parabola is

y = mx + \(\frac{a}{m}\)

Q(x_{1}, y_{1}) is the foot of the perpendicular

∴ y_{1} = mx_{1} + \(\frac{a}{m}\) ……….. (1)

Slope of SQ = \(\frac{y_{1}}{x_{1}-a}\)

SQ and PQ are perpendicular

Substituting in (1) we get

Locus of Q (x_{1}, y_{1}) is x = 0 i.e., the tangent at the vertex of the parabola.

Question 10.

Show that the tangent at one extremity of a focal chord of a parabola is parallel to the normal at the other extremity.

Solution:

P(t_{1}), Q(t_{2}) are the ends of a focai chord.

Slope of PS = Slope of PQ

Equation of the tangent at P(t_{1}) is

t_{1}y = x + at²_{1}

Slope of the tangent at P = \(\frac{1}{t_{1}}\) ………… (2)

Equation of the normal at Q(t_{2}) is

y + xt_{2} = 2at_{2} + at³_{2}

Slope of the normal at Q = -t_{2} ………… (3)

Frorn (1), (2)„ (3) we get slope of the tangent at P = slope of normal at Q

Slope of the tangent at P is parallel to the normal at Q.

III.

Question 1.

If the normal at the point t, on the parabola y² = 4ax meets it again at point t² then prove that t_{1}t_{2} + t_{1}² + 2 = 0.

Solution:

Equation of normal is

Equation of the line (i) again meets parabola at (at²_{2}, 2at_{2})

∴ 2at_{2} – 2at_{1} = t_{1}(at²_{2} – at²_{1})

–\(\frac{2}{t_{1}}\) = t_{1} + t_{2} ⇒ -2 = t²_{1} + t_{1}t_{2}

⇒ t²_{1} + t_{1}t_{2} + 2 = 0

Question 2.

From an external point P tangents are drawn to the parabola y² = 4ax and these tangents make angles θ_{1}, θ_{2} with its axis such that cot θ_{1} + cot θ_{2} is a constant ‘a’ show that P lies on a horizontal line.

Solution:

Equation of any tangent to the parabola is

y = mx + \(\frac{a}{m}\)

This tangent passes through P(x_{1}, y_{1})

y_{1} = mx_{1} + \(\frac{a}{m}\) ⇒ my_{1} = m²x_{1} + a

m²x_{1} – my_{1} + a = 0

Suppose m_{1}, m_{2} are the roots of the equation

m_{1} + m_{2} = \(\frac{y_{1}}{x_{1}}\), m_{1}m_{2} = \(\frac{a}{x_{1}}\)

Given cot θ_{1} + cot θ_{2} = a

Focus of P(x_{1}; y_{1}) is y = a² which is a horizontal line.

Question 3.

Show that the common tangent to the circle 2x² + 2y² = a² and the parabola 4ax intersect at the focus of the parabola y² = -4ax.

Solution:

Given circle is 2x² + 2y² = a²

with centre = (0, 0); Radius = \(\frac{a}{\sqrt{2}}\)

Given parabola is y² = 4ax a

Let y = mx + \(\frac{a}{m}\) be required tangent

But it touches 2x² + 2y² = a²

⇒ Perpendicular distance from (0, 0) = radius

Hence the focus of tangent of the parabola is y² = – 4ax.

Question 4.

The sum of the ordinates of two points on y² = 4ax is equal to the sum of the ordinates of two other points on the same curve. Show that the chord joining the first two points is parallel to the chord joining the other two points.

Solution:

Equation of the parabola is y² = 4ax

Equation of the chord joining P(t_{1}) and Q(t_{2}) is (t_{1} + t_{2})y = 2x + 2 at_{1}t_{2}

Slope of PQ = \(\frac{2}{t_{1}+t_{2}}\) ………… (1)

Equation of the chord joining R(t_{3}) and S(t_{4}) is (t_{3} + t_{4}) y = 2x + 2at_{3}t_{4}

Slope of RS = \(\frac{2}{t_{3}+t_{4}}\) ………… (2)

Given 2at_{1} + 2at_{2} = 2at_{3} + 2at_{4}

i.e., 2a(t_{1} + t_{2}) = 2a(t_{3} + t_{4})

t_{1} + t_{2} = t_{3} + t_{4}

From (1), (2), (3) we get slope of PQ = Slope of Rs

i.e., PQ and RS are parallel.

Question 5.

If a normal chord a point ‘t’ on the parabola y² = 4ax subtends a right angle at vertex, then prove that t = ± √2

Solution:

Equation of the parabola is y² = 4ax ………….. (1)

Equation of the normal at’t’ is

tx + y = 2at + at³

Homogenising (1) with the help of (2) combined equation of AQ, AR is

y² = \(\frac{4ax.(tx+y}{a(2t+t^{3})}\)

y²(2t + t³) = 4tx² + 4xy

4tx² + 4xy – (2t + t³)y² = 0

AQ, AR are perpendicular

Co-eff. of x² + Co.eff. of y² = 0

4t – 2t – t³ = 0

2t – t³ = 0

-t(t² – 2) = 0

t² – 2 = 0 ⇒ t² = 2

t = ± √2