Parabola

$x = 2t,\,y = {2 \over 3}$

$t \to 1\,\,\,A \equiv \left( {2,{1 \over 3}} \right)$

Given conic is ${x^2} = 12y \Rightarrow S \equiv (0,3)$

Let $B \equiv (0,\beta )$

Given $SA\, \bot \,BA$

$\left( {{{{1 \over 3}} \over {2 - 3}}} \right)\left( {{{\beta - {1 \over 3}} \over { - 2}}} \right) = - 1$

$ \Rightarrow \left( {\beta - {1 \over 3}} \right){1 \over 3} = - 2$

$ \Rightarrow \beta = {1 \over 3}\left( {{{ - 17} \over 3}} \right)$

$\mathop {Ordinate\,of\,centroid}\limits_{(as\,t \to 1)} = k = {{\beta + {1 \over 3} + 3} \over 3}$

$ = {{{{ - 17} \over 9} + {{10} \over 3}} \over 3} = {{13} \over {18}}$

According to the question (Let P(x, y))

$2x - y{{dx} \over {dy}} = 0$

($\because$ equation of tangent at $P:y - y = {{dy} \over {dx}}(y - x)$)

$\therefore$ $2{{dy} \over {y}} = {{dx} \over x}$

$ \Rightarrow 2\ln y = \ln x + \ln c$

$ \Rightarrow {y^2} = cx$

$\because$ this curve passes through (3, 3)

$\therefore$ c = 3

$\therefore$ required parabola ${y^2} = 3x$ and L.R = 3

For tangent to parabola $y=x^{2}$ at $(2,4)$

$ \left.\frac{d y}{d x}\right|_{(2,4)}=4 $

Equation of tangent is $ y-4=4(x-2) $

$\Rightarrow 4 x-y-4=0$

Family of circle can be given by

$(x-2)^{2}+(y-4)^{2}+\lambda(4 x-y-4)=0$

As it passes through $(0,6)$

$2^{2}+2^{2}+\lambda(-10)=0$

$\Rightarrow \lambda=\frac{4}{5}$

Equation of circle is

$ \begin{aligned} &(x-2)^{2}+(y-4)^{2}+\frac{4}{5}(4 x-y-4)=0 \\\\ &\Rightarrow \left(x^{2}+y^{2}-4 x-8 y+20\right)+\left(\frac{16}{5} x-\frac{4}{5} y-\frac{16}{5}\right)=0 \\\\ &A=-4+\frac{16}{5}, C=20-\frac{16}{5} \end{aligned} $

So, $A+C=16$

$\left( {y - {3 \over 4}} \right) = {\left( {x - {1 \over 2}} \right)^2}$ .... (1)

For $x = - {1 \over 2}$

$y - {3 \over 4} = 1 \Rightarrow y = {7 \over 4} \Rightarrow P\left( { - {1 \over 2},{7 \over 4}} \right)$

Now, $y' = 2\left( {x - {1 \over 2}} \right)$

At $x = - {1 \over 2}$

$ \Rightarrow {m_T} = - 2,{m_N} = {1 \over 2}$

Equation of normal is

$y - {7 \over 4} = {1 \over 2}\left( {x + {1 \over 2}} \right)$

$y = {x \over 2} + 2$

Now put y in equation (1)

${x \over 2} + 2 - {3 \over 4} = {\left( {x - {1 \over 2}} \right)^2}$

$ \Rightarrow x = 2$ & $ - {1 \over 2}$

$\Rightarrow$ Q(2, 3)

Now, ${(PQ)^2} = {{125} \over {16}}$

Option (b)

V $\to$ Vertex

F $\to$ focus

VF = S $-$ R

So, latus rectum = 4(S $-$ R)

Locus is directrix of parabola

x $-$ 3 + 4 = 0 $\Rightarrow$ x + 1 = 0.

Given, parabola

${y^2} = 8x$ ...... (i)

Equation of tangent at $P(2, - 4)$ is

$ - 4y = 4(x + 2)$

or, $x + y + 2 = 0$ ..... (ii)

and Equation of normal to the parabola is

$x - y + C = 0$

$\therefore$ Normal passes through $(2, - 4)$

$\therefore$ $C = - 6$

Normal : $x - y = 6$ ..... (iii)

Equation of directrix of parabola

$x = - 2$ ..... (iv)

Point of intersection of tangent and normal with directrix are $x = - 2$ at $A( - 2,0)$ and $B( - 2, - 8)$ respectively.

$Q(a,b)$ and $P(2, - 4)$ are given and AQBP is a square.

Mid-point of AB = Mid-point of PQ

$ \Rightarrow ( - 2, - 4) = \left( {{{a + 2} \over 2},{{b - 4} \over 2}} \right) \Rightarrow a = - 6,b = - 4$

$ \Rightarrow 2a + b = - 16$

Clearly RS is latus-rectum

$\because$ VF = 2 = a

$\therefore$ RS = 4a = 8

Now OF = 2a = 4

$\Rightarrow$ Area of triangle ORS = 16

Given, parabola $y = 4{x^2} + 1$


Let R(a, b) be mid-point of line joining point P and Q where PQ is perpendicular to line y = x.

Let coordinates of P be P(x, y), Q(q, q) and R(a, b) then,

$a = {{x + q} \over 2}$ and $b = {{y + q} \over 2}$

Now, slope of line y = x is m₁ = 1

Slope of line PQ be

${{b - q} \over {a - q}} = {m_2}$ (say)

$\because$ Line y = x and PQ are perpendicular to each other,

m₁ . m₂ = $-$1

$ \Rightarrow {{b - q} \over {a - q}} = - 1 \Rightarrow b - q = q - a$

$ \Rightarrow q = {{b + a} \over 2}$

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

$ \Rightarrow x = {{4a - b - a} \over 2} = {{3a - b} \over 2}$

and $b = {{y + q} \over 2} = {{y + \left( {{{b + a} \over 2}} \right)} \over 2} = {{2y + b + a} \over 4}$

$ \Rightarrow y = {{3b - a} \over 2}$

Put (x, y) in equation of parabola as P(x, y) is variable point on parabola

${{3b - a} \over 2} = 4{\left( {{{3a - b} \over 2}} \right)^2} + 1$

${{(3b - a)} \over 2} = {(3a - b)^2} + 1$

$ \Rightarrow (3b - a) = 2{(3a - b)^2} + 2$

Replace (a, b) as (x, y) $\Rightarrow$ (3y $-$ x) = 2(3x $-$ y)² + 2

or $2{(3x - y)^2} + (x - 3y) + 2 = 0$

Tangent at P : y(2) = 2(1/2) (x + 2)

$\Rightarrow$ 2y = x + 2

$\therefore$ Q = ($-$2, 0)

Normal at P : y $-$ 2 = $-$ ${{(2)} \over {2.{1 \over 2}}}(x - 2)$

$\Rightarrow$ y $-$ 2 = $-$ 2(x $-$ 2)

$\Rightarrow$ y = 6 $-$ 2x

$\therefore$ Solving with y² = 2x $\Rightarrow$ R$\left( {{9 \over 2} - 3} \right)$

$\therefore$ Ar ($\Delta$PQR) = ${1 \over 2}\left| {\matrix{ 2 & 2 & 1 \cr { - 2} & 1 & 1 \cr {{9 \over 2}} & 3 & { - 1} \cr } } \right|$

$ = {{25} \over 2}$ sq. units.

Parabola y² = 4x $-$ 20

Tangent at P(6, 2) will be

$2y = 4\left( {{{x + 6} \over 2}} \right) - 20$

2y = 2x + 12 $-$ 20

2y = 2x $-$ 8

y = x $-$ 4

x $-$ y $-$ 4 = 0 ....... (1)

This is also tangent to ellipse ${{{x^2}} \over 2} + {{{y^2}} \over b} = 1$

Apply c² = a²m² + b²

($-$4)² = (2)(1) + b

b = 14

Image of y² = 4x w.r.t. y = x is x² = 4y

tangent from (2, 1)

xx₁ = 2(y + y₁)

2x = 2(y + 1)

x = y + 1

Let the equation of the normal is

y = mx $-$ 2am $-$ am³

here 4a = 2 $ \Rightarrow $ a = ${1 \over 2}$

y = mx $-$ m $-$ ${1 \over 2}$m³

It passing through A(a, 0) then

0 = am $-$ m $-$ ${1 \over 2}$m³

m = 0, a $-$ 1 $-$ ${1 \over 2}$m² = 0

m² = 2(a $-$ 1) > 0

$ \Rightarrow $ a > 1

Shortest distance must be along common normal



Let a point on curve has x coordinate = h

Then y coordinate :

h² = 2y

$ \Rightarrow $ y = ${{{h^2}} \over 2}$

So point is = (h, ${{{h^2}} \over 2}$)

m₁ (slope of line x $-$ y = 1) = 1

$ \Rightarrow $ slope of perpendicular line = $-$1

Slope of the perpendicular line on the curve x² = 2y,

${m_2} = {{2x} \over 2} = x \Rightarrow {m_2} = h $

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

$ - {1 \over h}$ = $-$1 $ \Rightarrow $ h = 1

So point is $\left( {1,{1 \over 2}} \right)$

$D = \left| {{{1 - {1 \over 2} - 1} \over {\sqrt {1 + 1} }}} \right| = {1 \over {2\sqrt 2 }}$

Equation of tangent : $y = mx + {3 \over {2m}}$

${m_T} = {1 \over 2}$ ($\because$ perpendicular to line $2x + y = 1$)

$\therefore$ tangent is : $y = {x \over 2} + 3$

$ \Rightarrow x - 2y + 6 = 0$

Given, curve y = x² + 4

and, line y = 4x $-$ 1

Here, y = x² + 4


$\therefore$ ${{dy} \over {dx}} = 2x$ ..... (i)

and y = 4x $-$ 1

${{dy} \over {dx}} = 4$ ..... (ii)

Let the required point be P(x₁, y₁).

$\therefore$ ${\left. {{{dy} \over {dx}}} \right|_P} = 2{x_1}$ ..... (iii)

$\because$ Slopes will be equal.

$\therefore$ 2x₁ = 4 [from Eqs. (ii) and (iii)]

$ \Rightarrow {x_1} = {4 \over 2} = 2$

Now, the given point P(x₁, y₁) lies on curve y = x² + 4,

$\therefore$ y₁ = x$_1^2$ + 4

$\Rightarrow$ y₁ = 2² + 4 = 8

Hence, required coordinate of P = (2, 8)

Given, equation of parabola $\Rightarrow$ y² = 4ax

Focus = S(a, 0)

Let any point on the parabola be P(at², 2at).


and let the mid-point of PS be M(h, k).

$\therefore$ $h{{a{t^2} + a} \over 2};k = {{2at + 0} \over 2}$

$ \Rightarrow {t^2} = {{2h - a} \over a};t = {k \over a}$

$ \Rightarrow {t^2} = {{{k^2}} \over {{a^2}}}$

Now, ${{2h - a} \over a} = {{{k^2}} \over {{a^2}}}$

$ \Rightarrow 2h - a = {{{k^2}} \over a} \Rightarrow {k^2} = a(2h - a)$

$\therefore$ Locus of (h, k) is ${y^2} = a(2x - a)$

${y^2} = 2a\left( {x - {a \over 2}} \right)$

$\therefore$ The directrix of this parabola is

$x - {a \over 2} = - {a \over 2} \Rightarrow x = 0$

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 $