Parabola

Equation of given parabola is

y² = 4$\lambda $x ....(i)

So the end point of the latus rectum of the parabola (i), P($\lambda $, 2$\lambda $) and the given ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, passes through point P($\lambda $, 2$\lambda $).

On differentiating the equation of parabola, w.r.t. 'x', we get

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

$ \therefore $ Slope of tangent to the parabola at point P is m₁ = 1

Similarly, on differentiating the equation of given ellipse, ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$, w.r.t.x, we get ${{2x} \over {{a^2}}} + {{2y} \over {{b^2}}}{{dy} \over {dx}} = 0 \Rightarrow {{dy} \over {dx}} = - {{{b^2}x} \over {{a^2}y}}$

$ \therefore $ Slope of tangent to the ellipse at point P is ${m_2} = - {{{b^2}} \over {2{a^2}}}$

$ \because $ It is given that the tangents are perpendicular to each other. So, ${m_1}{m_2} = - 1$

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

$ \Rightarrow {{{b^2}} \over {{a^2}}} = 2 \Rightarrow b = \sqrt 2 a$

$ \therefore $ Eccentricity of ellipse ${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$ will be

$e = \sqrt {1 - {{{a^2}} \over {{b^2}}}} = \sqrt {1 - {1 \over 2}} = {1 \over {\sqrt 2 }}$ {$ \because $ b > a}

Given equation of parabola, $y^2=16 x$ and all the vertices are of an equilateral triangle with one of them coincides with the vertex of parabola

Since, given parabola is symmetrical about $X$-axis

So, $\angle A O M=30^{\circ}$

In $\triangle A O M, \tan 30^{\circ}=\frac{A M}{O M}=\frac{8 t}{4 t^2}$

$ \Rightarrow \quad \frac{1}{\sqrt{3}}=\frac{2}{t} \Rightarrow t=2 \sqrt{3} $

∴ Coordinate of point $A$ lying on parabola is

$ \left(4(2 \sqrt{3})^2, 8(2 \sqrt{3})\right) \equiv(48,16 \sqrt{3}) $

∴ Distance

$ \begin{aligned} O A & =\sqrt{\left(x_A-x_0\right)^2+\left(y_A-y_0\right)^2} \\ & =\sqrt{(48-0)^2+(16 \sqrt{3}-0)^2} \\ O A & =32 \sqrt{3} \end{aligned} $

∴ Length of the side of equilateral triangle $=32 \sqrt{3}$.

Given, equation of normal $m x-y+c=0$

Parabola $y^2=16 x$

Comparing with general equation of parabola i.e.,

$ y^2=4 a x \Rightarrow a=4 $

Let the coordinates of point $P$ on parabola be $\left(4 t^2, 8 t\right)$

∵ Slope of tangent at point $P=\frac{1}{t}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

Also, slope of tangent

$ \begin{equation*} =-\frac{1}{\text { Slope of normal }}=-\frac{1}{m} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) \end{equation*} $

From Eqs. (i) and (ii), $\frac{1}{t}=-\frac{1}{m}$ or $t=-m$

or $t=-m$

∴ Coordinate of point is $P\left(4 m^2,-8 m\right)$.

Focus of parabola is $(a, 0)$ or $(4,0)$.

Given, focal distance $=40$

Using distance formula,

$ \begin{aligned} & \sqrt{\left(4 m^2-4\right)^2+(-8 m-0)^2}=40 \\ & \text { or } \sqrt{\left(m^2-1\right)^2+(-2 m)^2}=10 \\ & \text { or } \sqrt{\left(m^2-1\right)^2+4 m^2}=10 \end{aligned} $

$ \begin{aligned} & \text { or } \sqrt{\left(m^2+1\right)^2}=10 \\ & \text { or } m^2+1=10 \\ & \text { or } m^2=9 \\ & \text { or } m= \pm 3 \end{aligned} $

Therefore, coordinate of point $P$

$ \left(4 \times( \pm 3)^2,-8( \pm 3)\right) \equiv(36, \pm 24) $

∵ Point $P$ lies on line $m x-y+c=0$ then it must satisfy equation of line

$ \begin{aligned} & & ( \pm 3)(36)-( \pm 48)+c & =0 \\ & & \pm(3 \times 36+48)+c & =0 \\ \text { or } & & c & = \pm 132 \\ \text { or } & & |c| & =132 \end{aligned} $

$P Q$ is focal chord of parabola $y^2=4 x$ focus $S(1,0)$ and $P(4,4)$

We know that, Length of Semi-latus rectum is Harmonic mean of $P S$ and $S Q$

$ \begin{aligned} & \therefore \quad \frac{1}{P S}+\frac{1}{S Q}=1 \Rightarrow \frac{1}{S Q}=1-\frac{1}{P S} \\ & \frac{1}{S Q}=1-\frac{1}{\sqrt{9+16}}=1-\frac{1}{5}=\frac{4}{5} \Rightarrow S Q=\frac{5}{4} \end{aligned} $

$ \begin{gathered} x^2=4 a y \\ y=1+2 x \text { intersect the parabola } \\ P(0,1) P A=r_1, P B=-r_2 \\ \frac{x-0}{\frac{1}{\sqrt{5}}}=\frac{y-1}{\frac{2}{\sqrt{5}}}=r \end{gathered} $

$ \begin{aligned} \Rightarrow \quad x=\frac{r}{\sqrt{5}}, y & =\frac{2 r}{\sqrt{5}}+1 \\ x^2 & =4 a y \\ \Rightarrow \quad \frac{r^2}{5} & =4 a\left(\frac{2 r}{\sqrt{5}}+1\right) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad r^2-8 \sqrt{5} a r-20 a=0 \\ & \Rightarrow \quad-r_1 r_2=-20 a \end{aligned} $

$ \begin{aligned} & \Rightarrow & r_1 r_2=20 a \Rightarrow r_1-r_2 & =8 \sqrt{5} a \\ & & \left(r_1+r_2\right)^2-\left(r_1-r_2\right)^2 & =4 r_1 r_2 \\ & & 40-64 \times 5 a^2 & =80 a \\ & \Rightarrow & 8 a^2+2 a-1 & =0 \\ & & (4 a-1)(2 a+1) & =0 \\ & \Rightarrow & 4 a & =1 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { Given, } y=\frac{h^3}{3} x^2+\frac{h^2}{2} x-h+\frac{3}{4 h^3} \\ & 12 h^3 y=4 h^6 x^2+6 h^5 x-12 h^4+9 \\ & 4 h^6 x^2+6 h^5 x=12 h^3 y+12 h^4-9 \\ & \begin{aligned} \left(x^2+\frac{3}{2 h} x\right) & =\frac{12 h^4}{4 h^6}\left(y+\frac{12 h^4}{12 h^3}-\frac{3}{14 h^3}\right) \\ \left(x^2+\frac{3}{2 h^2}+\right. & \left.\frac{3}{16 h^2}\right) \\ & =\frac{3}{h^3}\left(y+h-\frac{3}{4 h^3}\right)+\frac{9}{16 h^2} \end{aligned} \end{aligned} \end{aligned} $

$ \left(x+\frac{3}{2 h}\right)^2=\frac{3}{h^3}\left(y+\frac{19 h}{16}-\frac{3}{4 h^3}\right) $

Here, $a=\frac{3}{4 h^3}$

∴ Equation of directrix is

$ \begin{aligned} & y+\frac{19 h}{16}-\frac{3}{4 h^3}=\frac{-3}{4 h^3} \Rightarrow y=-\frac{19 h}{16} \\ & \therefore \quad k=-\frac{19 h}{16} \Rightarrow \frac{k}{h}=-\frac{19}{16} \\ & \therefore \quad k: h=-19: 16 \end{aligned} $

Given equation of parabola

$ x^2=108 y(4 a=108 \Rightarrow a=27) $

Equation of tangent, $y=m x-27 m^2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

And equation of parabola

$ y^2=32 x(4 a=32 \Rightarrow a=8) $

Equation of tangent, $y=m x+\frac{8}{m}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

Since Eqs. (i) and (ii) are identical

$ \begin{aligned} & \therefore \quad m x-27 m^2=m x+\frac{8}{m} \\ & \Rightarrow \quad-27 m^2=\frac{8}{m} \Rightarrow m^3=\frac{-8}{27} \Rightarrow m=\frac{-2}{3} \end{aligned} $

∴ Required equation of tangent

$ \begin{array}{cc} & y=\left(\frac{-2}{3}\right) x-27\left(\frac{-2}{3}\right)^2 \\ \Rightarrow & y=\frac{-2 x}{3}-12 \\ \Rightarrow & 3 y=-2 x-36 \\ \Rightarrow & 2 x+3 y+36=0 \end{array} $

We have, equation of parabola is

$ \begin{array}{cr} & y^2+2 x+2 y-3=0 \\ \Rightarrow & (y+1)^2-1+2 x-3=0 \\ \Rightarrow & (y+1)^2=4-2 x \\ \Rightarrow & (y+1)^2=-2(x-2) \\ \therefore & \text { Vertex }=(2,-1) \end{array} $

Axis : $y+1=0$

Focus: $(-a+2,-1)=\left(-\frac{1}{2}+2,-1\right)$

                                         $ =\left(\frac{3}{2},-1\right) $

Directrix : $x-2=\frac{1}{2} \Rightarrow x=\frac{5}{2}$

$ \Rightarrow 2 x-5=0 $

We have, $y^2=4 x$

We know that, the centroid of the triangle formed by the conormal points on a parabola lies on its axis and its coordinates are $\left(\frac{2}{3}(h-2 a), 0\right)$, where $(h, k)$ is the point through which all normals passes. $\quad[\therefore a=1]$ ∴ Coordinate of centroid $=\left(\frac{2}{3}(5-2 \times 1), 0\right)=(2,0)$

Given equation of parabola is

x = 4y $-$ y²

$\Rightarrow$ y² $-$ 4y + 4 = $-$ x + 4

$\Rightarrow$ (y $-$ 2)² = $-$ (x $-$ 4)

a = ${{1} \over 4}$

Focus = $\left( {{{15} \over 4},2} \right)$

Parabola cuts Y-axis of (0, 0) and (0, 4)

Now, equation of tangent at (0, 0) is

x = 4y ...... (i)

and equation of tangent at (0, 4) is

x + 4y = 16 .... (ii)

Solving Eqs. (i) and (ii), we get

x = 8, y = 2

$\therefore$ Intersection point of tangent is P(8, 2).

Distance from point P to focus S

Distance = $\sqrt {{{\left( {8 - {{15} \over 4}} \right)}^2} + {{(2 - 2)}^2}} $

$ = 8 - {{15} \over 4} = {{17} \over 4}$

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}$