Parabola

Circle passes through A(0, 1) and B(2, 4).

y = x²

$ \Rightarrow $ ${\left. {{{dy} \over {dx}}} \right|_B}$ = 4

tangent at (2,4) is

(y – 4) = 4(x – 2)

4x – y – 4 = 0

Equation of circle

(x - 2)² + (y–4)² + $\lambda $(4x–y - 4) = 0

Passing through (0,1)

$ \therefore $ 4 + 9 + $\lambda $(–5) = 0

$ \Rightarrow $ $\lambda $ = ${{13} \over 5}$

$ \therefore $ Circle is

x²– 4x + 4 + y² – 8y + 16 + ${{13} \over 5}$[4x - y - 4] = 0

$ \Rightarrow $ x² + y² + $\left( {{{52} \over 5} - 4} \right)$x - $\left( {8 + {{13} \over 5}} \right)$y + 20 - ${{{52} \over 5}}$ = 0

$ \Rightarrow $ x² + y² + ${{{32} \over 5}x - {{53} \over 5}y}$ + ${{48} \over 5}$ = 0

$ \therefore $ Centre is $\left( {{{ - 16} \over 5},{{53} \over {10}}} \right)$

L₁ : y² = 4(x + 1)

Equation of tangent y = m(x + 1) + ${1 \over m}$ ...(1)

L₂ : y² = 8(x + 2)

Equation of tangent y = m'(x + 2) + ${2 \over {m'}}$

$ \Rightarrow $ y = m'x + 2$\left( {m' + {1 \over {m'}}} \right)$ ....(2)

Since lines intersect at right angles

$ \therefore $ mm' = -1

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

Putting it in equation (2)

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

$ \Rightarrow $ y = $ - {1 \over m}x - 2\left( {m + {1 \over m}} \right)$ ....(3)

From equation (1) and (3)

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

$ \Rightarrow $ $\left( {m + {1 \over m}} \right)x + 3\left( {m + {1 \over m}} \right)$ = 0

$ \therefore $ x + 3 = 0

$y = mx + {1 \over m}$ (tangent at y² = 4x)

y = mx – m² (tangent at x² = 4y)

${1 \over m} = - {m^2}$ (for common tangent)

m³ = – 1

$ \Rightarrow $ m = - 1

$ \therefore $ Equation of tangent

y = –x –1

x + y + 1 = 0

This line touches circle whose center at (0, 0),

$ \therefore $ apply p( Distance from center of the circle of the line ) = r ( Radius of the circle )

c = $\left| {{{0 + 0 + 1} \over {\sqrt 2 }}} \right| = {1 \over {\sqrt 2 }}$

For parabola y² = 4x,

Length of latus rectum = 4a = 4

C₁C₂ = 2(C₁A)

C₁A = $\sqrt {{{(2\sqrt 5 )}^2} - {2^2}} = 4$

$ \therefore $ C₁C₂ = 2(4) = 8

Let P = (3t² , 6t) ; N = (3t² , 0)

M = (3t² , 3t)

Let point Q = (h, 3t) which lies on the parabola y² = 12x

$ \therefore $ 9t² = 12h

$ \Rightarrow $ h = ${{3{t^2}} \over 4}$

Equation of NQ

y - 0 = ${{3t} \over {{{3{t^2}} \over 4} - 3{t^2}}}\left( {x - 3{t^2}} \right)$

$ \Rightarrow $ y = ${{ - 4} \over {3t}}\left( {x - 3{t^2}} \right)$

For y-intercept of NQ, x = 0

$ \therefore $ y = 4t

Given y-intercept of the line NQ is ${4 \over 3}$

$ \therefore $ 4t = ${4 \over 3}$

$ \Rightarrow $ t = ${1 \over 3}$

$ \therefore $ MQ = ${{9{t^2}} \over 4}$ = ${1 \over 4}$

PN = 6t = 2

Let A = (2t² , 4t)

B = (2t² , -4t) (by symmetry as equilateral triangle)


tan 30^o = $\frac{4t}{2t^{2}} $ = $\frac{2}{t} $

$ \Rightarrow $ t = 2$\sqrt{3} $

AB = 8t = 16$\sqrt{3} $

Area of equilateral = $\frac{1}{2} \times 16\sqrt{3} \times 24$

= $192\sqrt 3 $

Given parabola y² = 8x

$ \therefore $ a = 2

Let one end of focal chord is A(at² , 2at) = $\left( {{1 \over 2}, - 2} \right)$

$ \therefore $ 2at = -2

$ \Rightarrow $ t = $ - {1 \over 2}$

Other end of focal chord will be B$\left( {{a \over {{t^2}}}, - {{2a} \over t}} \right)$ $ \equiv $ (8, 8)

$ \therefore $ Equation of tangent at B is

8y = 4(x + 8)

$ \Rightarrow $ x – 2y + 8 = 0

Take point P(2t, t² ) on parabola x² = 4y

h = ${{2t + 0} \over 3}$ and k = ${{{t^2} - 2} \over 3}$

$ \Rightarrow $ t = ${{3h} \over 2}$ and 3k + 2 = t²

$ \therefore $ 3k + 2 = ${{9{h^2}} \over 4}$

$ \Rightarrow $ 9x² – 12y = 8

Given y = mx + 4 is tangent to both the parabolas.

$ \therefore $ Applying condition of tangent for y² = 4x, we get

${1 \over m}$ = 4

$ \Rightarrow $ m = ${1 \over 4}$

For x² = 2by line y = ${x \over 4}$ + 4 is tangent

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

$ \Rightarrow $ x² - ${{bx} \over 2}$ - 8b = 0

For tangent to the parabola Discriminant = 0

$ \Rightarrow $ ${{{b^2}} \over 4}$ + 32b = 0

$ \Rightarrow $ b = -128

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

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 $

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$

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 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}$

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 - $2.$

The locus of perpendicular tangents is directrix

i.e., $x=-1$

Vertex of a parabola is the mid point of focus and the point

where directrix meets the axis of the parabola.

Here focus is $O\left( {0,0} \right)$ and directrix meets the axis at $B\left( {2,0} \right)$

$\therefore$ Vertex of the parabola is $(1,0)$

Parabola ${y^2} = 8x$



We know that the locus of point of intersection of two perpendicular tangents to a

parabola is its directrix. Point must be on the directrix of parabola

as equation of directrix $x + 2 = 0 \Rightarrow x = - 2$

Hence the point is $\left( { - 2,0} \right)$

Given parabola is $y = {{{a^3}{x^2}} \over 3} + {{{a^2}x} \over 2} - 2a$

$ \Rightarrow y = {{{a^3}} \over 3}\left( {{x^3} + {3 \over {2a}} + x + {9 \over {16{a^2}}}} \right) - {{3a} \over {16}} - 2a$

$ \Rightarrow y + {{35a} \over {16}} = {{{a^3}} \over 3}{\left( {x + {3 \over {4a}}} \right)^2}$

$\therefore$ Vertex of parabola is $\left( {{{ - 3} \over {4a}},{{ - 35a} \over {16}}} \right)$

To find locus of this vertex,

$x = {{ - 3} \over {4a}}\,\,$ and $\,\,y = {{ - 35a} \over {16}}$

$ \Rightarrow a = {{ - 3} \over {4x}}\,\,$ and $a = - {{16y} \over {35}}$

$ \Rightarrow {{ - 3} \over {4x}} = {{ - 16y} \over {35}} \Rightarrow 64xy = 105$

$ \Rightarrow xy = {{105} \over {64}}$ which is the required locus.

$P = \left( {1,0} \right)\,\,Q = \left( {h,k} \right)$ Such that ${k^2} = 8h$

Let $\left( {\alpha ,\beta } \right)$ be the midpoint of $PQ$

$\alpha = {{h + 1} \over 2},\,\,\,\beta = {{k + 0} \over 2}$

$ \therefore $ $2\alpha - 1 = h\,\,\,\,\,\,2\beta = k.$

${\left( {2\beta } \right)^2} = 8\left( {2\alpha - 1} \right) \Rightarrow {\beta ^2} = 4\alpha - 2$

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

Solving equations of parabolas

${y^2} = 4ax$ and ${x^2} = 4ay$

we get $(0,0)$ and $(4a, 4a)$

Substituting in the given equation of line

$2bx+3cy+4d=0,$

we get $d=0$

and $2b+3c=0$ $ \Rightarrow {d^2} + {\left( {2b + 3c} \right)^2} = 0$

Equation of the normal to a parabola ${y^2} = 4bx$ at point

$\left( {bt_1^2,2b{t_1}} \right)$ is $y = - {t_1}x + 2b{t_1} + bt_1^3$

As given, it also passes through $\left( {bt_2^2,2b{t_2}} \right)$ then

$2b{t_2} = {t_1}bt_2^2 + 2b{t_1} + bt_1^3$

$2{t_2} - 2{t_1} = - {t_1}\left( {t_2^2 - t_1^2} \right)$

$ = - {t_1}\left( {{t_2} + {t_1}} \right)\left( {{t_2} - {t_1}} \right)$

$ \Rightarrow 2 = - {t_1}\left( {{t_2} + {t_1}} \right)$

$ \Rightarrow {t_2} + {t_1} = - {2 \over {{t_1}}}$

$ \Rightarrow {t_2} = - {t_1} - {2 \over {{t_1}}}$