Parabola

Let the equation of tangent to parabola

y² = 16x is y = mx + ${4 \over m}$ ...... (1)

It is given that tangent to xy = -4 .........(2)

Solving (1) and (2) we get

$x\left( {mx + {4 \over m}} \right) + 4 = 0$

$ \Rightarrow m{x^2} + {4 \over m}x + 4 = 0$

Now for tangent D = 0 $ \Rightarrow {{16} \over {{m^2}}} - 16m = 0$

$ \Rightarrow {m^3} = 1$ $ \Rightarrow $ m = 1

Now putting value of m in Equation (1)

y = x + 4 or x – y + 4 = 0

Let the coordinates of C be (h, k)

So the chord of contact of C w.r.t y = (x-2)² - 1

$ \Rightarrow $ y = x² - 4x + 3 is T = 0

$y+k \over 2$ = xh + 3 - 2(x+h)

$ \Rightarrow $ 2(h-2)x - y = -(6-4h-k)

On comparing it with x - y = 3

2 (h -2) = 1 $ \Rightarrow $ h = $5 \over 2$

6 - 4h - k = -3 $ \Rightarrow $ k = -1

$ \therefore $ C = ($5 \over 2$, -1)

Let equation of common tangent is y = mx + ${3 \over m}$

$ \therefore $ ${\left( {{3 \over m}} \right)^2}$ = 1, m² - 8

$ \Rightarrow {m^4} - 8{m^2} - 9 = 0$

$ \Rightarrow {m^2} = 9 \Rightarrow m = \pm 3$

$ \therefore $ equation of common tangents are y = 3x + 1 & y = -3x - 1

$ \therefore $ ${{PS} \over {PS}} = {{3 + {1 \over 3}} \over { - {1 \over 3} + 3}} = {5 \over 4}$

Tangent to the curve y² = 4$\sqrt 2$x is y = mx + ${{\sqrt 2 } \over m}$

It is tangent to the circle x² + ^y2 = 1

$ \therefore $ $\left| {{{\sqrt 2 /m} \over {\sqrt {1 + {m^2}} }}} \right| = 1 \Rightarrow m = \pm 1$

$ \therefore $ tangent are y = x + $\sqrt 2$ & y = – x – $\sqrt 2$

Compare with y = – ax + c

$ \Rightarrow a = \pm 1 $ & $ c = \pm \sqrt 2 $

Equation of tangent to the parabola y² = 4x at P(1, 2),

T = 0

2y = $4\left( {{{x + 1} \over 2}} \right)$

$ \Rightarrow $ y = x + 1

Equation of normal of the tangent at point P(1, 2)

y - 2 = (-1)(x - 1)

$ \Rightarrow $ y - 2 = - x + 1

$ \Rightarrow $ x + y - 3 = 0

This normal also passes through the center (h, r) of the circle.

$ \therefore $ h + k - 3 = 0

$ \Rightarrow $ h = 3 - r

So center is (3 - r, r)

From picture you can see,

PC = r

$ \Rightarrow $ (PC)² = r²

$ \Rightarrow $ (3 - r - 1)² + (r - 2)² = r²

$ \Rightarrow $ 4 + r² - 4r + r² + 4 - 4r = r²

$ \Rightarrow $ r² - 8r + 8 = 0

$ \therefore $ r = ${{8 \pm \sqrt {64 - 32} } \over 2}$

$ \Rightarrow $ r = ${{8 \pm \sqrt {32} } \over 2}$

$ \Rightarrow $ r = ${{8 \pm 4\sqrt 2 } \over 2}$

$ \therefore $ r = 4 + ${2\sqrt 2 }$ and 4 - ${2\sqrt 2 }$

If r = 4 + ${2\sqrt 2 }$ then center of the circle is

(-1 - ${2\sqrt 2 }$, 4 + ${2\sqrt 2 }$). From the diagram you can see both the x coordinate and y coordinate of the circle should be positive but here x coordinate is negative.

So possible value of radius r = 4 - ${2\sqrt 2 }$

Then area of the circle

= $\pi $r²

= $\pi $(4 - ${2\sqrt 2 }$)²

= $\pi $(16 + 8 - ${16\sqrt 2 }$)

= $8\pi \left( {3 - 2\sqrt 2 } \right)$

For this parabola y² = 16x,

a = 4

Here PQ is focal cord.

Let P(at₁², 2at₁) and Q(at₂², 2at₂).

Given P(1, 4),

$ \therefore $ at₁² = 1

$ \Rightarrow $ 4t₁² = 1

$ \Rightarrow $ t₁² = ${1 \over 4}$

$ \Rightarrow $ t₁ = ${1 \over 2}$

In parabola if the parameter of one end point of the focal cord is t₁ then parameter of the other end point t₂ = $ - {1 \over {{t_1}}}$

Here parameter for point Q t₂ = - 2

$ \therefore $ Length of focal cord

|PQ| = a${\left( {{t_1} - {t_2}} \right)^2}$ = 4${\left( {{1 \over 2} + 2} \right)^2}$ = 25

Parabola y² = 4x and circle x² + y² = 5 intersect with each other.

So, x² + 4x = 5

$ \Rightarrow $ x² + 5x – x – 5 = 0

$ \Rightarrow $ x(x + 5) –1(x + 5) = 0

x = 1, –5

Intersection point in 1^st quadrant is = (1, 2)

Equation of tangent to y² = 4x at (1, 2) is

y(2) = 2 (x + 1)

$ \Rightarrow $ y = x + 1 .....(1)

By checking each options, you can see

point $\left( { {3 \over 4},{7 \over 4}} \right)$ lies on equation (1).

Slope of the line y = x is 1.

$ \therefore $ The shortest distance between line y = x and parabola y² = x – 2 is the distance between line y = x and tangent of parabola having slope 1.

Slope of tangent at point P on the parabola y² = x – 2 is,

2y${{dy} \over {dx}}$ = 1

$ \Rightarrow $ ${{dy} \over {dx}}$ = ${1 \over {2y}}$ = slope

$ \therefore $ ${1 \over {2y}}$ = 1

$ \Rightarrow $ y = ${1 \over 2}$

$ \therefore $ x coordinate of point P is,

${1 \over 4}$ = x – 2

$ \Rightarrow $ x = ${9 \over 4}$

Shortest distance = length of perpendicular from P on line x – y = 0

= ${{\left| {{9 \over 4} - {1 \over 2}} \right|} \over {\sqrt {{1^2} + {1^2}} }}$ = ${7 \over {4\sqrt 2 }}$

x² = 8y

$ \Rightarrow $  ${{dy} \over {dx}} = {x \over 4} = \tan \theta $

$ \therefore $  x₁ = 4tan$\theta $

y₁ = 2 tan² $\theta $

Equation of tangent :-

y $-$ 2tan²$\theta $ = tan$\theta $ (x $-$ 4tan $\theta $)

$ \Rightarrow $  x = y cot $\theta $ + 2 tan $\theta $

y² = 4x

2yy' = 4

y ' = ${1 \over t} = 2,$   $t = {1 \over 2}$

Area = ${1 \over 2}\left| {\matrix{ {{1 \over 4}} & 1 & 1 \cr 9 & 6 & 1 \cr 4 & { - 4} & 1 \cr } } \right| = {{225} \over 4}$

f (a) = 2a(12 $-$ a)²

f '(a) = 2(12 $-$ 3a²)

Maximum at a = 2

maximum area = f(2) = 32

Vertex is (a², 0)

y² $=$ $-$(x $-$ a²) and x $=$ 0 $ \Rightarrow $ (0, $ \pm $ 2a)

Area of triangle is $ = {1 \over 2}.$4a.(a²) = 250

$ \Rightarrow $  a³ = 125 or a = 5

x² = 4y

x $-$ $\sqrt 2 $y + 4$\sqrt 2 $ = 0

Solving together we get

x² = 4$\left( {{{x + 4\sqrt 2 } \over {\sqrt 2 }}} \right)$

$\sqrt 2 $x² + 4x + 16$\sqrt 2 $

$\sqrt 2 $x² $-$ 4x $-$ 16$\sqrt 2 $ = 0

x₁ + x₂ = 2$\sqrt 2 $; x₁x₂ = ${{ - 16\sqrt 2 } \over {\sqrt 2 }}$ = $-$ 16

Similarly,

($\sqrt 2 $y $-$ 4$\sqrt 2 $)² = 4y

2y² + 32 $-$ 16y = 4y



$\ell $_AB = $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $

$ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 64 + {{\left( {10} \right)}^2} - 4\left( {16} \right)} $

$ = \sqrt {8 + 64 + 100 - 64} $

$ = \sqrt {108} = 6\sqrt 3 $

Normal to the two given curves are

y = m(x – c) – 2bm – bm³,

y = mx – 4am – 2am³

If they have a common normal, then

(c + 2b)m + bm³ = 4am + 2am³

$ \Rightarrow $ (4a – c – 2b) m = (b – 2a)m³

$ \Rightarrow $ (4a – c – 2b) = (b – 2a)m²

$ \Rightarrow $ m² = ${c \over {2a - b}} - 2$ $>$ 0

By checking all options we found (A) is right option.

Note : We get that all the options are correct for m = 0 i.e., when common normal is x-axis. But may be in question they want common normal other than x – axis, hence answer is (A).

$\Delta ABC = {1 \over 2}\left| {\matrix{ 4 & { - 4} & 1 \cr 9 & 6 & 1 \cr {{t^2}} & {2t} & 1 \cr } } \right|$

D = 60 + 10t $-$ 10t²

${{d\Delta } \over {dt}} = 0 \Rightarrow t = {1 \over 2}$

${{{d^2}\Delta } \over {d{t^2}}} = - 20 < 0$

$ \therefore $  max at $t = {1 \over 2}$

max area $\Delta = 65 - {5 \over 2}$

$ = {{125} \over 2} = 31{1 \over 4}$

So the equation of the parabola,

${\left( {y - 0} \right)^2} = 4.a\left( {x - 2} \right)$

$ \Rightarrow $  y² = 4.2 (x $-$ 2)

$ \Rightarrow $  y² = 8 (x $-$ 2)

By checking each options you can see. point (8, 6) does not lie on the parabola.

Angle between the curves is the acute angle between the tangents at the point of intersection.

y = 10 $-$ x² (for curve 1)

and y = 2 + x² (for curve 2)

$ \therefore $  10 $-$ x² = 2 + x²

$ \Rightarrow $  2x² = 8

$ \Rightarrow $  x² = 4

$ \Rightarrow $  x = 2, $-$ 2

$ \therefore $  points of intersection (2, 6) and ($-$ 2, 6)

${{dy} \over {dx}}$ for curve 1 = $-$ 2x

$ \therefore $  Slope(m₁) of curve 1 is = $-$ 2(2) = $-$ 4

${{dy} \over {dx}}$ for curve 2 = 2x

$ \therefore $  slope (m₂) of curve 2 = 2 $ \times $ 2 = 4

$ \therefore $  tan$\theta $ = $\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$

= $\left| {{{ - 4 - 4} \over {1 + \left( { - 16} \right)}}} \right|$

= ${8 \over {15}}$

We know,

Equation of tangent to the parabola y² = 4ax is,

y = mx + ${a \over m}$

$ \therefore $  Equation of tangent to the parabola y² = 4x is,

y = mx + ${1 \over m}$

$ \Rightarrow $  m²x $-$ ym + 1 = 0

This tangent is also the tangent to the circle x² + y² $-$ 6x = 0

So, the perpendicular distance from the center of the circle to the tangent is equal to the radius of the circle.

Here center is at (3, 0) of the circle and radius = 3

$ \therefore $  $\left| {{{3{m^2} + 1} \over {\sqrt {{m^4} + {m^2}} }}} \right| = 3$

$ \Rightarrow $  (3m² + 1)² = 9(m⁴ + m²)

$ \Rightarrow $  9m⁴ + 6m² + 1 = 9m⁴ + 9m²

$ \Rightarrow $  3m² = 1

$ \Rightarrow $  m = $ \pm $ ${1 \over {\sqrt 3 }}$

So, possible tangents are

y = ${1 \over {\sqrt 3 }}$x + $\sqrt 3 $

$ \Rightarrow $  $\sqrt 3 $y = x + 3

or   y = $-$ ${x \over {\sqrt 3 }}$ $-$ $\sqrt 3 $

$ \Rightarrow $  $\sqrt 3 y$ = $-$ x $-$ 3

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear,

$ \therefore $ $\left| {\matrix{ 0 & 0 & 1 \cr 3 & 4 & 1 \cr h & k & 1 \cr } } \right| = 0 \Rightarrow 4h = 3k$ ....(i)

and MN is the length of diameter of circle C₃, so

MN = $3 + \sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} + 4$

= $3 + 5 + 4 = 12$

So, radius of circle C₃, r = 6 ......(ii)

Since, the circle C₃ touches C₁ at M and C₂ at N, so

|C₁ C₃| = |r $-$ 3|

$ \Rightarrow $ $\sqrt {{h^2} + {k^2}} = 3$

$ \Rightarrow $ h² + k² = 9 .....(iii)

From Eqs. (i) and (iii), we get

${h^2} + {{16{h^2}} \over 9} = 9 \Rightarrow 25{h^2} = 81$

$ \Rightarrow h = + {9 \over 5}$ and $k = + {{12} \over 5}$

So, $2h + k = {{18} \over 5} + {{12} \over 5} = 6$

Now, equation common chord XY of circles C₁ and C₂ is

C₁ $-$ C₂ = 0

$ \Rightarrow $ 6x + 8y = 18

$ \Rightarrow $ 3x + 4y = 9 ....(iv)

Now, PY² = GY² $-$ GP²

= $9 - {{81} \over {25}} = {{144} \over {25}}$

$ \Rightarrow $ PY = ${{12} \over 5}$

$ \because $ $XY = 2PY = 2 \times {{12} \over 5} = {{24} \over 5}$

Similarly, equation of ZW is 3x + 4y = 9.

Now, So, length of perpendicular from C₃ to

$ZW = {{\left| {3\left( {{9 \over 5}} \right) + 4\left( {{{12} \over 5}} \right) - 9} \right|} \over 5} = {6 \over 5}$

So, $P{W^2} = {C_3}{W^2} - {C_3}{P^2} = 36 - {{36} \over {25}} = {{864} \over {25}}$

{$ \because $ C₃W = r = 6}

$ \Rightarrow $ $PW = {{12\sqrt 6 } \over 5}$

$ \because $ $ZW = 2PW = {{24\sqrt 6 } \over 5}$

$ \therefore $ ${{length\,of\,ZW} \over {length\,of\,XY}} = \sqrt 6 $

Now, area of

$\Delta MZN = {1 \over 2}(MN)(PZ) = {1 \over 2} \times (12)\left( {{1 \over 2}WZ} \right)$

{$ \because $ MN = 12}

= $3WZ = 3{{24\sqrt 6 } \over 5} = {{72\sqrt 6 } \over 5}$

and area of $\Delta $ZMW = ${1 \over 2}$(ZW) (MP)

= ${1 \over 2}\left( {{{24\sqrt 6 } \over 5}} \right)(MG + GP)$

$ = {{12\sqrt 6 } \over 5}\left( {3 + {9 \over 5}} \right)$

{$ \because $ MG = 3 and GP = ${{9 \over 5}}$}

$ = {{12\sqrt 6 } \over 5}\left( {{{24} \over 5}} \right)$

= ${{288\sqrt 6 } \over 5}$

$ \therefore $ ${{Area\,of\,\Delta MZN} \over {Area\,of\,\Delta ZMW}} = {{{{72\sqrt 6 } \over 5}} \over {{{288\sqrt 6 } \over 5}}} = {5 \over 4}$

$ \because $ Common tangent of circles C₁ and C₃ is C₁ $-$ C₃ = 0

$ \Rightarrow $ $({x^2} + {y^2} - 9) - \left[ {{{\left( {x - {9 \over 5}} \right)}^2} + {{\left( {y - {{12} \over 5}} \right)}^2} - 36} \right] = 0$

$ \Rightarrow $ ${{18} \over 5}x + {{24} \over 5}y + 18 = 0$

$ \Rightarrow $ 3x + 4y + 15 = 0 ......(v)

$ \because $ Tangent (v) is also touches the parabola x² = 8$\alpha $y,

$ \therefore $ $ - 2\alpha {\left( { - {3 \over 4}} \right)^2} = - {{15} \over 4}$

$ \Rightarrow $ $\alpha = {{10} \over 3}$

So combination (iv), (S) is only incorrect.

Hence, option (b) is correct.

It is given that, the centres of circles C₁, C₂ and C₃ are co-linear,

$ \therefore $ $\left| {\matrix{ 0 & 0 & 1 \cr 3 & 4 & 1 \cr h & k & 1 \cr } } \right| = 0 \Rightarrow 4h = 3k$ ....(i)

and MN is the length of diameter of circle C₃, so

MN = $3 + \sqrt {{{(3 - 0)}^2} + {{(4 - 0)}^2}} + 4$

= $3 + 5 + 4 = 12$

So, radius of circle C₃, r = 6 ......(ii)

Since, the circle C₃ touches C₁ at M and C₂ at N, so

|C₁ C₃| = |r $-$ 3|

$ \Rightarrow $ $\sqrt {{h^2} + {k^2}} = 3$

$ \Rightarrow $ h² + k² = 9 .....(iii)

From Eqs. (i) and (iii), we get

${h^2} + {{16{h^2}} \over 9} = 9 \Rightarrow 25{h^2} = 81$

$ \Rightarrow h = + {9 \over 5}$ and $k = + {{12} \over 5}$

So, $2h + k = {{18} \over 5} + {{12} \over 5} = 6$

Now, equation common chord XY of circles C₁ and C₂ is

C₁ $-$ C₂ = 0

$ \Rightarrow $ 6x + 8y = 18

$ \Rightarrow $ 3x + 4y = 9 ....(iv)

Now, PY² = GY² $-$ GP²

= $9 - {{81} \over {25}} = {{144} \over {25}}$

$ \Rightarrow $ PY = ${{12} \over 5}$

$ \because $ $XY = 2PY = 2 \times {{12} \over 5} = {{24} \over 5}$

Similarly, equation of ZW is 3x + 4y = 9.

Now, So, length of perpendicular from C₃ to

$ZW = {{\left| {3\left( {{9 \over 5}} \right) + 4\left( {{{12} \over 5}} \right) - 9} \right|} \over 5} = {6 \over 5}$

So, $P{W^2} = {C_3}{W^2} - {C_3}{P^2} = 36 - {{36} \over {25}} = {{864} \over {25}}$

{$ \because $ C₃W = r = 6}

$ \Rightarrow $ $PW = {{12\sqrt 6 } \over 5}$

$ \because $ $ZW = 2PW = {{24\sqrt 6 } \over 5}$

$ \therefore $ ${{length\,of\,ZW} \over {length\,of\,XY}} = \sqrt 6 $

So, combination (ii), Q is only correct.

Hence, option (c) is correct.

Let P(2t, t²) be any point on the parabola.

Center of the given circle C = ($-$ g, $-$f) = ($-$3, 0)

For PC to be minimum, it must be the normal to the parabola at P.

Slope of line PC = ${{{y_2} - {y_1}} \over {{x_2} - {x_1}}}$ = ${{{t^2} - 0} \over {2t + 3}}$

Also, slope of tangent to parabola at P = ${{dy} \over {dx}}$ = ${x \over 2}$ = t

$ \therefore $ Slope of normal = ${{ - 1} \over t}$

$ \therefore $ ${{{t^2} - 0} \over {2t + 3}}$ = ${{ - 1} \over t}$

$ \Rightarrow $ t³ + 2t + 3 = 0

$ \Rightarrow $ (t+1) (t² $-$ t + 3) = 0

$\therefore\,\,\,$ Real roots of above equation is

t = $-$ 1

Coordinate of P = (2t, t²) = ($-$2, 1)

Slope of tangent to parabola at P = t = $-$ 1

Therefore, equation of tangent is :

(y $-$ 1) = ($-$ 1) (x + 2)

$ \Rightarrow $ x + y + 1 = 0

As equation of tangent PA at (x₁, y₁) on the parabola y² = 4ax,

yy₁ = 2a (x + x₁)

here (x₁, y₁) = ( 16, 16)

y . 16 = 2.4 (x + 16)

$ \Rightarrow $ 2y = x + 16 .....(1)

At pont A value of y = 0

putting y = 0 in equation (1) we get,

0 = x + 16

$ \Rightarrow $ x = $-$ 16

$\therefore\,\,\,$ Coordinate of point A = ($-$ 16, 0)

Slope of line P A :

As$\,\,\,\,$ 2y = x + 16

$\therefore\,\,\,$ y = ${1 \over 2}\,$ x + 8

$\therefore\,\,\,$ Slope (m) = ${1 \over 2}\,$

Let slope of perpendicular line PB passing through point p(16, 16) = m'

$\therefore\,\,\,$ m m' = $-$ 1

$ \Rightarrow $ ${1 \over 2}$ $ \times $ m' = $-$ 1

$ \Rightarrow $ ' = $-$ 2

As Equation of normal PB, when slope is m,

y = mx $-$ 2am $-$am³

Here m = m' = $-$ 2 and a = 4

$\therefore\,\,\,$ y = $-$2x $-$ 2(4) ($-$2) $-$ 4 . ($-$2)³

$ \Rightarrow $ y = $-$ 2x + 16 + 32

$ \Rightarrow $ y = $-$ 2x + 48 ..... (2)

At point B, y = 0

puttig y = 0 at equation (2) we get,

0 = $-$ 2x + 48

$ \Rightarrow $x = 24

$\therefore\,\,\,$ Coordinate of point B = (24, 0)



A circle is passing through point P, A and B, and C is the center of the circle.

So, AC and BC are the radius.

Then AC = BC, So C is the middle point of line AB.

$\therefore\,\,\,$ C = $\left( {{{24 - 16} \over 2},{{0 + 0} \over 2}} \right)$ = (4, 0)

$\angle $CPB = $\theta $ and we have to find tan $\theta $.

Slope of line PC = ${{16 - 0} \over {16 - 4}}$ = ${4 \over 3}$ =m₁

and we know slope of line PB = $-$2 = m₂

$\therefore\,\,\,$ tan $\theta $ = $\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$

$ \Rightarrow $ tan $\theta $ = $\left| {{{{4 \over 3} + 2} \over {1 - \left( {{4 \over 3}} \right)\left( 2 \right)}}} \right|$

$ \Rightarrow $ tan $\theta $ = $\left| {{{{{10} \over 3}} \over { - {5 \over 3}}}} \right|$

$ \Rightarrow $ tan $\theta $ = $\left| { - 2} \right|$

$ \Rightarrow $ tan $\theta $ = 2

Equation of the chord of contact PQ is given by : T=0

or T $ \equiv $ yy₁ $-$ 4(x + x₁), where (x₁, y₁) $ \equiv $ ($-$8, 0)

$\therefore\,\,\,$Equation becomes : x = 8

& chord of contact is x = 8

$\therefore\,\,\,$ Coordinates of point P and Q are (8, 8) and (8, $-$ 8)

and focus of the parabola is F (2, 0)

$\therefore\,\,\,$ Area of triangle PQF = ${1 \over 2}$ $ \times $ (8 $-$ 2) $ \times $ (8 + 8) = 48 sq. units

As origin is the only common point to x-axis and y-axis, so origin is the common vertex

Let the equation of two of parabolas be y² = 4ax and x² = 4by

Now latus rectum of both parabolas = 3

$\therefore\,\,\,$ 4a = 4b = 3

$ \Rightarrow $$\,\,\,$ a = b = ${3 \over 4}$

$\therefore\,\,\,$ Two parabolas are y² = 3x and x² = 3y

Suppose y = mx + c is in the common tangent.

$\therefore\,\,\,$ y² = 3x $ \Rightarrow $ (mx + c)² = 3x $ \Rightarrow $ m²x² + (2mc $-$ 3) x + c² = 0

As, the tangent touches at one point only

So, b² $-$ 4ac = 0

$ \Rightarrow $ (2mc $-$ 3)² $-$ 4m²c² = 0

$ \Rightarrow $ 4m²c² + 9 $-$ 12mc $-$ 4m²c² = 0

$ \Rightarrow $ c = ${9 \over {12m}}$ = ${3 \over {4m}}$      . . . .(i)

$\therefore\,\,\,$ x² = 3y $ \Rightarrow $ x² = 3 (mx + c ) $ \Rightarrow $ x² $-$ 3mx $-$ 3c = 0

Again, b² $-$ 4ac = 0

$ \Rightarrow $ 9m² $-$ 4(1) ($-$3c) = 0

$ \Rightarrow $ 9m² = $-$ 12c       . . . . .(ii)

From (i) and (ii)

m² = ${{ - 4c} \over 3}$ = ${{ - 4} \over 3}$ $\left( {{3 \over {4m}}} \right)$

$ \Rightarrow $$\,\,\,$ m³ = $-$ 1 $ \Rightarrow $ m = $-$ 1 $ \Rightarrow $ c = ${{ - 3} \over 4}$

Hence, y = mx + c = $-$ x $-$ ${3 \over 4}$

$ \Rightarrow $ 4 (x + y) + 3 = 0

c = $-$ 29m $-$ 9m³

a = 2

Given (at² $-$ a)² + 4a²t² = 64

$ \Rightarrow $   (a(t² + 1)) = 8

$ \Rightarrow $   t² + 1 = 4 $ \Rightarrow $ t² = 3

$ \Rightarrow $   t = $\sqrt 3 $

$ \therefore $   c = 2at(2 + t²)

= $2\sqrt 3 \left( 5 \right)$

$\left| c \right|$ = 10$\sqrt 3 $

Tangent to x² + y² = 4 is

y = mx $ \pm $ 2$\sqrt {1 + {m^2}} $

Also, x² = 4y

x² = 4mx + 8$\sqrt {1 + {m^2}} $

  or  x² = 4mx $-$ 8$\sqrt {1 + {m^2}} $

For D = 0

we have; 16m² + 4.8$\sqrt {1 + {m^2}} $ = 0

$ \Rightarrow $   m² + 2$\sqrt {1 + {m^2}} $ = 0

$ \Rightarrow $    m² = $-$ 2$\sqrt {1 + {m^2}} $

$ \Rightarrow $   m⁴ = 4 + 4m²

$ \Rightarrow $   m⁴ $-$ 4m² $-$ 4 = 0

$ \Rightarrow $    m² = ${{4 \pm \sqrt {16 + 16} } \over 2}$

$ \Rightarrow $   m² = ${{4 \pm 4\sqrt 2 } \over 2}$

$ \Rightarrow $   m² = 2 + 2$\sqrt 2 $

The tangent is y = x + 8.

Satisfying (8, 16) in y² = 4ax, we have

16² = 4a . 8. $\therefore$ a = 8

On comparing y = mx + ${a \over m}$, we have m = 1

The point of contact is $\left( {{a \over {{m^2}}},{{2a} \over m}} \right) = (8,16)$

So its verified.

Given : The equation of parabola is y² = 16x

Equation chord is

2x + y = p ..... (1)

Equation of chord with middle point (h, k) is given as

$ky - 16\left( {{{x + h} \over 2}} \right) = {k^2} - 16h$

$ \Rightarrow ky - 8(x + h) = {k^2} - 16h$

$ \Rightarrow ky - 8x - 8h = {k^2} - 16h$

$ \Rightarrow ky - 8x - 8h - {k^2} + 16h = 0$

$ \Rightarrow - 8x + ky + 8h - {k^2} = 0$

$ \Rightarrow 8x - ky = 8h - {k^2}$ ..... (2)

Comparing above equation with equation of chord, we get

$2x + y = p$

Dividing Eq. (2) by 4, we get

${{8x} \over 4} - {{ky} \over 4} = {{8h} \over 4} - {{{k^2}} \over 4}$

$2x - {{ky} \over 4} = 2h - {{{k^2}} \over 4}$

On comparing, we get

$ \Rightarrow {{ - k} \over 4} = 1$ and $p = 2h - {{{k^2}} \over 4}$

$ \Rightarrow k = - 4$ and $p = 2h - {{{{( - 4)}^2}} \over 4} = 2h - 4$

$ \Rightarrow p = 2h - 4 \Rightarrow 2h - p = 4$

Only p = 2 and h = 3 satisfies this equation. Therefore, p = 2, h = 3 and k = $-$4.

t₁ = $-$ t $-$ ${2 \over t}$

$t_1^2$ = t² + ${4 \over {{t^2}}}$ + 4

t² + ${4 \over {{t^2}}}$ $ \ge $ 2$\sqrt {{t^2}.{4 \over {{t^2}}}} = 4$

Minimum value of $t_1^2$ = 8

Minimum distance $ \Rightarrow $ perpendicular distance

$E{q^n}$ of normal at $p\left( {2{t^2},\,4t} \right)$

$y = - tx + 4t + 2{t^3}$

It passes through $C\left( {0, - 6} \right) \Rightarrow {t^3} + 2t + 3 = 0$

$ \Rightarrow t = - 1$



Center of new circle $ = P\left( {2t{}^2,4t} \right) = P\left( {2, - 4} \right)$

Radius $ = PC = \sqrt {{{\left( {2 - 0} \right)}^2} + {{\left( { - 4 + 6} \right)}^2}} = 2\sqrt 2 $

$\therefore$ Equation of the circle is

${\left( {x - 2} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {2\sqrt 2 } \right)^2}$

$ \Rightarrow {x^2} + y{}^2 - 4x + 8y + 12 = 0$

Tangent to y² = 4x at (t², 2t) is

y(2t) = 2(x + t²)

$\Rightarrow$ yt = x + t² ...... (i)

Equation of normal at P(t², 2t) is

y + tx = 2t + t³

Since, normal at P passes through centre of circle S(2, 8).

$\therefore$ 8 + 2t = 2t + t³ $\Rightarrow$ t = 2, i.e., P(4, 4)

$\therefore$ $SP = \sqrt {{{(4 - 2)}^2} + {{(4 - 8)}^2}} = 2\sqrt 5 $

$\therefore$ Option (a) is correct.

Also, SQ = 2

$\therefore$ PQ = SP $-$ SQ = 2$\sqrt5$ $-$ 2

Thus, ${{SQ} \over {QP}} = {1 \over {\sqrt 5 - 1}} = {{\sqrt 5 + 1} \over 4}$

$\therefore$ Option (b) is incorrect.

Now, x-intercept of normal is

x = 2 + 2² = 6

$\therefore$ Option (c) is correct.

Slope of tangent $ = {1 \over t} = {1 \over 2}$

$\therefore$ Option (d) is correct.

In case of option (A)

x² + y² = 3 and x² = 2y

Solving we get $P(\sqrt 2 ,1)$

Equation of tangent at P on the circle

$x\sqrt 2 + y = 3$ $\therefore$ $\tan \alpha = - \sqrt 2 $

Again, ${{{Q_2}{R_2}} \over {{Q_2}M}} = \sin \left( {\alpha - {\pi \over 2}} \right)$

or, ${{2\sqrt 3 } \over {{Q_2}M}} = - \cos \alpha = - {1 \over {\sqrt 3 }}$ [$\because$ $\tan \alpha = - \sqrt 2 $]

or, ${Q_2}M = 6$

Similarly, ${Q_3}M = 6$

$\therefore$ ${Q_2}{Q_3} = 12$

In case of option (B)

${{M{R_2}} \over {{Q_2}{R_2}}} = \tan \left( {\alpha - {\pi \over 2}} \right)$

or, ${{M{R_2}} \over {2\sqrt 3 }} = - \tan \alpha = \sqrt 2 $

or, $M{R_2} = 2\sqrt 6 $

Similarly, $M{R_3} = 2\sqrt 6 $ $\therefore$ ${R_2}{R_3} = 4\sqrt 6 $

In case of option (C)

OP is the perpendicular drawn from O to R₂R₃

$\therefore$ area of $\Delta O{R_2}{R_3} = {1 \over 2} \times OP \times {R_2}{R_3} = {1 \over 2} \times \sqrt 3 \times 4\sqrt 6 =6\sqrt 2 $

In case of option (D)

PT is perpendicular drawn from P to Q₂Q₃

$\therefore$ area of $\Delta P{Q_2}{Q_3} = {1 \over 2} \times PT \times {Q_2}{Q_3} = {1 \over 2} \times \sqrt 2 \times 12 = 6\sqrt 2 $

Therefore, (A), (B), (C) are correct options.

Let the coordinates of Q and P be (x₁, y₁) and (h, k) respectively.

$\because$ Q lies on x² = 8y,

$\therefore$ x$_1^2$ = 8y ....... (1)

Again, P divides OQ internally in the ratio 1 : 3.

$\therefore$ $h = {{{x_1} + 0} \over 4} = {{{x_1}} \over 4}$ or x₁ = 4h and

$k = {{{y_1} + 0} \over 4} = {{{y_1}} \over 4}$ or y₁ = 4k

Now putting x₁ and y₁ in (1) we get,

16h² = 32k or, h² = 2k

$\therefore$ the locus of P is given by, x² = 2y.

Let $P\left( {{{t_1^2} \over 2},{t_1}} \right)$ and $Q\left( {{{t_2^2} \over 2},{t_2}} \right)$ be two distinct points on the parabola ${y^2} = 2x$.

The circle with PQ as diameter passes through the vertex O(0, 0) of the parabola.

Clearly, PO $\bot$ OQ

So, slope of PO $\times$ slope of OQ = $-$1

or, ${{{t_1} - 0} \over {{{t_1^2} \over 2} - 0}} \times {{{t_2} - 0} \over {{{t_2^2} \over 2} - 0}} = - 1$

or, ${2 \over {{t_1}}} \times {2 \over {{t_2}}} = - 1$

or, ${t_1}{t_2} = - 4$

By question, area of $\Delta OPQ = 3\sqrt 2 $

or, ${1 \over 2}\left| {\matrix{ 0 & 0 & 1 \cr {{{t_1^2} \over 2}} & {{t_1}} & 1 \cr {{{t_2^2} \over 2}} & {{t_2}} & 1 \cr } } \right| = 3\sqrt 2 $

or, ${1 \over 2}\left| {{{t_1^2{t_2}} \over 2} - {{{t_1}t_2^2} \over 2}} \right| = 3\sqrt 2 $

or, $\left| {{t_1}{t_2}} \right|\left| {{t_1} - {t_2}} \right| = 12\sqrt 2 $

or, $\left| {{t_1} + {4 \over {{t_1}}}} \right| = 3\sqrt 2 $ [$\because$ ${t_1}{t_2} = - 4$]

or, ${t_1} + {4 \over {{t_1}}} = 3\sqrt 2 $ [$\because$ P lies in first quadrant]

or, $t_1^2 - 3\sqrt 2 {t_1} + 4 = 0$

or, ${t_1} = {{3\sqrt 2 \pm \sqrt {18 - 4 \times 1 \times 4} } \over 2}$

$ = {{3\sqrt 2 \pm \sqrt 2 } \over 2}$

$ = 2\sqrt 2 ,\sqrt 2 $

$\therefore$ coordinates of $P = (4,2\sqrt 2 )$ or $(1,\sqrt 2 )$.

Therefore, (A) and (D) are correct options.

Let tangent to ${y^2} = 4x$ be $y = mx + {1 \over m}$

Since this is also tangent to ${x^2} = - 32y$

$\therefore$ ${x^2} = - 32\left( {mx + {1 \over m}} \right)$

$ \Rightarrow {x^2} + 32mx + {{32} \over m} = 0$

Now, $D=0$

${\left( {32} \right)^2} - 4\left( {{{32} \over m}} \right) = 0$

$ \Rightarrow {m^3} = {4 \over {32}} \Rightarrow m = {1 \over 2}$

We know that on the parabola $y^2=4 a x$, the equation of tangent at $p=\left(a t^2, 2 a t\right)$ is $t y=x+ a t^2$ and Normal at $\mathrm{S}\left(a s^2, 2 a s\right)$ is $y+s x=2 a s+a s^3$ Let tangent at P and Normal at Q intersect at $\left(x_0, y_0\right)$, then

$\begin{aligned} t y_0 & =x_0+a t^2 \quad \text{.... (i)}\\ y_0+s x_0 & =2 a s+a s^3 \quad \text{... (ii)} \end{aligned}$

From equation (i) and (ii)

$\begin{aligned} & x_0=t y_0-a t^2=\frac{-y_0+2 a s+a s^3}{s} \\ \Rightarrow & \text { st } y_0-\text { at.t.s }=-y_0+2 a s+a s^3 \end{aligned}$

Given, $s t=1$

$\begin{aligned} & \therefore \quad y_0-a t=-y_0+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad 2 y_0=a t+\frac{2 a}{t}+\frac{a}{t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^4+2 t^2+1\right)}{2 t^3} \\ & \Rightarrow \quad y_0=\frac{a\left(t^2+1\right)^2}{2 t^3} \end{aligned}$

Hint:

On the parabola $y^2=4 a x$

(i) The equation of tangent at $\mathrm{P}\left(a t^2, 2 a t\right)$ is $ty=x+a t^2$.

(ii) The equation of normal at $\mathrm{S}\left(a s^2, 2 a s\right)$ is $y+s x=2 a s+a s^3$.

Given, four points $\mathrm{P}\left(a t^2, 2 a t\right), \mathrm{Q}, \mathrm{R}\left(a r^2, 2 a r\right)$ and $S\left(a s^2, 2 a s\right)$ be distinct points on the parabola $y^2=4 a x$.

Also given PQ is the focal chord of parabola

$\begin{aligned} & y^2=4 a x \\ & \therefore \quad \mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right) \end{aligned}$

Given, $\mathrm{K}=(2 a, 0)$ and PK is parallel to QR

$\therefore$ Slope of line PK $=$ Slope of line QR

$\begin{aligned} & \frac{2 a t-0}{a t^2-2 a}=\frac{2 a r+\frac{2 a}{t}}{a r^2-\frac{a}{t^2}} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{r+\frac{1}{t}}{r^2-\frac{1}{t^2}} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{r+\frac{1}{t}}{\left(r+\frac{1}{t}\right)\left(r-\frac{1}{t}\right)} \\ \Rightarrow & \frac{t}{t^2-2}=\frac{1}{r-\frac{1}{t}} \\ \Rightarrow & t r-1=t^2-2 \\ \Rightarrow & \quad r=\frac{t^2-1}{t} \end{aligned}$

Hint:

(i) If PQ is a focal chord of parabola $y^2=4 a x$ and $\mathrm{P}=\left(a t^2, 2 a t\right)$, then $\mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

(ii) If two lines are parallel, then their slopes are equal.

(iii) The slope of line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{y_2-y_1}{x_2-x_1}$.

Let common tangent be

$y = mx + {{\sqrt 5 } \over m}$

Since, perpendicular distance from center of the circle to

the common tangent is equal to radius of the circle,

therefore ${{{{\sqrt 5 } \over m}} \over {\sqrt {1 + {m^2}} }} = \sqrt {{5 \over 2}} $

On squaring both the side, we get

${m^2}\left( {1 + {m^2}} \right) = 2$

$ \Rightarrow {m^4} + {m^2} - 2 = 0$

$ \Rightarrow \left( {{m^2} + 2} \right)\left( {{m^2} - 1} \right) = 0$

$ \Rightarrow m = \pm 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,$ ( as $m \ne \pm \sqrt 2 $ )

$y = \pm \left( {x + \sqrt 5 } \right),$ both statements are correct as $m = \pm 1$

satisfies the given equation of statement - $2.$

Given, a parabola $y^2=16 x$

Let $\mathrm{F}\left(x_0, y_0\right)=\left(4 t^2, 8 t\right)$

$\Rightarrow \quad x_0=4 t^2, y_0=8 t$

Equation of tangent at F is $t y=x+4 t^2$

G is the point of intersection of tangent $t y=x +4 t^2$ and $Y$-axis

$\therefore \mathrm{G}=(0,4 t)$

Given, $\mathrm{G}=\left(0, y_1\right)$

$\therefore \quad y_1=4 t$

Let $\Delta$ be the area of $\Delta$ EGF

$\begin{aligned} & \Rightarrow \quad \Delta=\frac{1}{2} 4 t^2 \cdot\left(3-y_1\right) \\ & \Rightarrow \quad \Delta=2 t^2(3-4 t) \\ & \Rightarrow \quad \frac{d \Delta}{d t}=12 t-24 t^2=0 \text { at } t=0, \frac{1}{2} \\ & \Rightarrow \frac{d^2 \Delta}{d t^2}=12-48 t<0 \text { at } t=\frac{1}{2} \end{aligned}$

Hence, $\Delta$ is maximum at $t=\frac{1}{2}$

$\therefore \quad \Delta_{\max }=\frac{1}{2}, y_0=8 t=4, y_1=4 t=2$

Given, $y=m x+3$ intersect the parabola $y^2=16 x$ at $\mathrm{F}\left(4 t^2, 8 t\right)$

$\therefore 8 t=4 m t^2+3$

Put $t=\frac{1}{2}$

$\begin{aligned} & \Rightarrow \quad 4=m+3 \\ & \Rightarrow m=1 \end{aligned}$

Hints :

(i) $\frac{d y}{d x}=0$ at $x=\alpha$ and $\frac{d^2 y}{d x^2}<0$ at $x=\alpha$, then the function has point of local maxima at $x=a$.

(ii) The parametric point of the parabola $y^2= 4 a x$ is $\left(a t^2, 2 a t\right)$

(iii) The equation of tangent of parabola $y^2=4 a x$ at $\left(a t^2, 2 a t\right)$ is $t y=x+a t^2$.

Given, PQ be a focal chord of the parabola $y^2=4 a x$

Let $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$

Equation of tangent at P is $t y=x+a t^2\quad \text{... (i)}$

Equation of tangent at Q is $-\frac{y}{t}=x+\frac{a}{t^2}\quad \text{... (ii)}$

Let $R$ is the point of intersection of tangents at $P$ and $Q$.

On solving Equation (i) and (ii)

$\Rightarrow \mathrm{R}=\left(-a, a\left(t-\frac{1}{t}\right)\right)$

Given, the tangents of parabola $y^2=4 a x$ at P and $Q$ meet at a point lying on the line $y=2 x +a$.

$\therefore \mathrm{R}$ must be lies on the line $y=2 x+a$

$\begin{aligned} & \Rightarrow \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad t-\frac{1}{t}=-1 \quad \text{... (iii)} \end{aligned}$

Let the length of chord PQ is L

$\begin{aligned} & \Rightarrow \mathrm{L}=\sqrt{\left(a t^2-\frac{a}{t^2}\right)^2+\left(2 a t+\frac{2 a}{t}\right)^2} \\ & \Rightarrow \mathrm{L}=a \sqrt{\left(t-\frac{1}{t}\right)^2\left(t+\frac{1}{t}\right)^2+4\left(t+\frac{1}{t}\right)^2} \\ & \Rightarrow \mathrm{L}=a \sqrt{\left(t-\frac{1}{t}\right)^2\left(\left(t-\frac{1}{t}\right)^2+4\right)+4\left(\left(t-\frac{1}{t}\right)^2+4\right)} \\ & \Rightarrow \mathrm{L}=a \sqrt{1 \cdot(1+4)+4 \cdot(1+4)} \\ & \Rightarrow \mathrm{L}=5 a \end{aligned}$

Hints:

If $P Q$ is a focal chord of parabola $y^2=4 a x$, then $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}=\left(\frac{a}{t^2}, \frac{-2 a}{t}\right)$

Equation of tangent at P is $t y=x+a t^2\quad \text{.... (i)}$

Equation of tangent at Q is $\frac{-y}{t}=x+\frac{a}{t^2}\quad \text{... (ii)}$

On solving equations (i) and (ii)

$\Rightarrow \mathrm{R}=\left(-a, a\left(t-\frac{1}{t}\right)\right)$

Given, the tangents of parabola $y^2=4 a x$ at P and Q meet at a point lying on the line $y=2 x+a$

$\begin{aligned} & \Rightarrow \quad a\left(t-\frac{1}{t}\right)=-2 a+a \\ & \Rightarrow \quad\left(t-\frac{1}{t}\right)=-1 \quad \text{... (iii)} \end{aligned}$

$\Rightarrow$ Slope of line $\mathrm{OP}=\frac{2}{t}$ and Slope of line OQ $=-2 t$

Given, $\angle \mathrm{POQ}=\theta$

$\begin{aligned} & \Rightarrow \quad \tan \theta=\frac{\frac{2}{t}-(-2 t)}{1+\frac{2}{t}(-2 t)} \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3}\left(t+\frac{1}{t}\right) \\ & \Rightarrow \quad \tan \theta=\frac{-2}{3} \sqrt{\left(t-\frac{1}{t}\right)^2+4} \\ & \Rightarrow \quad \tan \theta=\frac{-2 \sqrt{5}}{3} \end{aligned}$

Hints:

(i) If slope of two lines are $m_1$ and $m_2$, then the angle of intersection between them is equal to $\tan ^{-1}\left(\frac{m_1-m_2}{1+m_1 m_2}\right)$

(ii) If PQ is a focal chord of parabola $y^2=4 a x$, then $\mathrm{P}=\left(a t^2, 2 a t\right)$ and $\mathrm{Q}=\left(\frac{a}{t^2},-\frac{2 a}{t}\right)$.

Now, ${y^2} = 4x$ and Q will lie on it.

$ \Rightarrow {(4k)^2} = 4 \times 4h$

$ \Rightarrow {k^2} = h$

$ \Rightarrow {y^2} = x$ (replacing h by x and k by y)

The equation of normal is

y = mx $-$ 2m $-$ m³

As (9, 6) lies on it, 6 = 9m $-$ 2m $-$ m³3 $-$ 7m + 6 = 0

The roots are m = 1, 2, $-$3. So the normal are

y = x $-$ 3, y = 2x $-$ 12, y = $-$3x + 33.

The locus of perpendicular tangents is directrix

i.e., $x=-1$

Let A $\equiv$ (t$_1^2$, 2t₁) and B $\equiv$ (t$_2^2$, 2t₂)

The centre of the circle = $\left( {{{t_1^2 + t_2^2} \over 2},{t_1} + {t_2}} \right)$

As the circle touches the x-axis thus ${t_1} + {t_2} = \pm \,r$

Slope of $AB = {2 \over {{t_1} + {t_2}}} = \pm \,{2 \over r}$