Parabola

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 $\frac{y_2-y_1}{x_2-x_1}$.

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.

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

The intersection point of $y = 0$ with first line is $B( - p,0)$.

The intersection point of $y=0$ with second line is $A( - q,0)$.

The intersection point of the two lines is $C(pq,(p + 1)(q + 1))$.

The altitude from C to AB is $x = pq$.

The altitude from B to AC is

$y = - {q \over {1 + q}}(x + p)$

Solving these two equations, we get $x = pq$ and $y = - pq$.

Hence, the locus of orthocentre is $x + y = 0$.

We have $G \equiv (h,k)$.

$ \Rightarrow h = {{2a + a{t^2}} \over 3},k = {{2at} \over 3}$

$ \Rightarrow \left( {{{3h - 2a} \over a}} \right) = {{9{k^2}} \over {4{a^2}}}$

Therefore, the required parabola is

${{9{y^2}} \over {4{a^2}}} = {{(3x - 2a)} \over a} = {3 \over a}\left( {x - {{2a} \over 3}} \right)$

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

Hence, the vertex $ \equiv \left( {{{2a} \over 3},0} \right)$; focus $ \equiv (a,0)$.

$\begin{aligned} & y=\frac{-x^{2}}{2}+x+1 \\ & 2 y=-x^{2}+2 x+2 \\ & 2 y=-\left(x^{2}-2 x+1\right)+3 \end{aligned}$

$(2 y-3)=-(x-1)^{2} \Rightarrow$ parabola is symmetric with respect to $x=1$

Also statement 1 is true statement 2 is correct explanation of statement 1.

The tangent to the curve $y=e^x$ drawn at the point ($c,e^c$) is given as:

$(y-e^c)=e^c(x-c)$ ..... (i)

The equation of the line joining the points

$(c-1),(e^{c-1})$ and $(c+1,e^{c+1})$

$y-e^{c-1}=\frac{e^c(e-e^{-1})}{2}(x-c+1)$ ..... (ii)

On subtracting (i) from (ii)

${e^c} - {e^{c - 1}} = {{{e^c}(e - {e^{ - 1}})} \over 2}(x - c + 1) - {e^c}(x - c)$

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

$x - c = {{{{(e - 1)}^2}} \over {2 - {{(e - 1)}^2}}} < 0$

$ \Rightarrow x < c$

We first obtain the coordinate of P & Q.

Towards this end we solve

${x^2} + {y^2} = 9$ ..... (i)

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

from (i) and (ii), we get

${x^2} + 8x - 9 = 0$

$(x + 9)(x - 1) = 0 \Rightarrow x = 1$ [$\therefore$ $x + 9 > 0$]

${y^2} = 8 \Rightarrow y = \, \pm 2\sqrt 2 $

Thus, coordinates of P are $(1,2\sqrt 2 )$ and Q are $(1, - 2\sqrt 2 )$ equation of tangents to the circle at P and Q respectively.

$x + 2\sqrt 2 y = 9$

$x - 2\sqrt 2 y = 9$

On solving these we get coordinates of R as (9, 0)

Equations of tangents to parabola at P and Q are

$2\sqrt 2 y = 4(x + 1)$

$ - 2\sqrt 2 y = 4(x + 1)$

These lines intersect each other at S($-1,0$)

Let $\Delta_1$ = area of $\Delta$PQS = $\frac{1}{2}$ (PQ)(ST)

$=\frac{1}{2}(2)(4\sqrt2)$

$=4\sqrt2$

And $\Delta_2$ = area of $\Delta$PQR = $\frac{1}{2}$(PQ)(TR)

$=\frac{1}{2}(4\sqrt2)8$

$=16\sqrt2$

$\Delta_1:\Delta_2=4\sqrt2:16\sqrt2$

= 1 : 4

Equation of perpendicular bisector of SR is x = 4

Midpoint of PR is (5, $\sqrt2$) ..... (i)

Slope of PR is ${{0 - 2\sqrt 2 } \over {9 - 1}} = {{ - \sqrt 2 } \over 4}$

$\therefore$ Equation of perpendicular bisector of PR is

$y - \sqrt 2 = {4 \over {\sqrt 2 }}(x - 5)$ ...... (ii)

Solving equation (i) and (ii) we get

$x = 4,y = - \sqrt 2 $

Circumcentre is (4, $-\sqrt2$)

Radius of circumcircle is = 3$\sqrt3$ units

We have

PR = RQ = 6$\sqrt2$

PQ = 4$\sqrt2$

$\Delta_2$ = area of $\Delta$PQR

= 16$\sqrt2$

$r = {{{\Delta _2}} \over s}$

$ = {{16\sqrt 2 } \over {{1 \over 2}(6\sqrt 2 + 6\sqrt 2 + 4\sqrt 2 )}} = 2$ units

Radius of incircle of $\Delta$PQR = 2 units

Axis of parabola along line $y=x$ distance of vertex from origin is $\sqrt{2}$. It implies parabola lies in first Quadrant.

Equation of directrix is line passing through the origin and perpendicular on $y=x$ i.e, $y=-x \Rightarrow x+y=0$

Now by definition of parabola. Parabola is a locus of a point which moves in such a way its distance from fixed point and from fixed line are equal where fixed point is called focus and fixed line is called directrix.

i.e, if $\mathrm{P}(x, y)$ be any point on parabola then Distance of P from directrix $=$ Distance of P from focus

$ \begin{aligned} & \Rightarrow & \left(\frac{x+y}{\sqrt{1^2+1^2}}\right)^2 & =(x-2)^2+(y-2)^2 \\ & \Rightarrow & (x+y)^2 & =2\left[x^2+y^2-4 x-4 y+8\right] \\ & \Rightarrow & x^2+y^2+2 x y & =2 x^2+2 y^2-8 x-8 y+16 \\ & \Rightarrow & x^2+y^2-2 x y & =8(x+y-2) \\ & \Rightarrow & (x-y)^2 & =8(x+y-2) \end{aligned} $

$ y^2=4 x $

normal on parabola

$ \begin{aligned} 2 y \frac{d y}{d x} & =4 \\ \frac{d y}{d x} & =\frac{2}{y} \\ \frac{-d x}{d y} & =\frac{-y}{2} \end{aligned} $

Equation of normal

$ \begin{aligned} y & =\left(\frac{-y}{2}\right)(x-3) \\ \Rightarrow 2 & =3-x \\ \Rightarrow x & =1 \text { or } y=0 \end{aligned} $

Points $(\mathrm{P}, \mathrm{Q}, \mathrm{R})$ are $(0,0)(1,2)$ and $(1,-2)$

Base $\mathrm{QR}=4$ and Height $=1$

(i) area of $\triangle \mathrm{PQR}=0.5 \times 4 \times 1$

(ii) circum radius $=\frac{a b c}{4 a}=\frac{20}{4 \times 2}=\frac{5}{2}$

(iii) centroid of $\triangle \mathrm{PQR}$

$ =\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right)\left(\frac{2}{3}, 0\right) $

(iv) circum centre: since, the triangle is an isosceles triangle and circum centre lies $\frac{5}{2}$ distance apart (radius)

From origin.

Therefore, circumcentre $\left(\frac{5}{2}, 0\right)$

Let’s say the common tangent line has the equation $y = m x + c$.

We want this line to touch both parabolas: $y = x^2$ and $y = -(x-2)^2$.

1. Tangent to $y = x^2$:

For the line to be a tangent, it must touch the curve at just one point. Set $y = x^2$ and $y = m x + c$ equal:

$x^2 = m x + c$

Bring all terms to one side:

$x^2 - m x - c = 0$

This quadratic has one solution (double root) if its discriminant is zero. The discriminant is $b^2 - 4ac$ in $ax^2 + bx + c = 0$.

Here, $a = 1$, $b = -m$, $c = -c$.

Set discriminant to zero:

$(-m)^2 - 4 \times 1 \times (-c) = 0$

$m^2 + 4c = 0$

So, $c = -\frac{m^2}{4}$

That means the tangent equation is:

$y = m x - \frac{m^2}{4}$

2. Same line tangent to $y = -(x-2)^2$:

The line must also touch the second parabola at one point. Set their equations equal:

$-(x-2)^2 = m x - \frac{m^2}{4}$

Expand $-(x-2)^2$:

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

So:

$-x^2 + 4x - 4 = m x - \frac{m^2}{4}$

Bring all terms to one side:

$x^2 + (m - 4)x + \left(4 - \frac{m^2}{4}\right) = 0$

Again, for a tangent, discriminant is zero:

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

Calculate this step-by-step:

$(m - 4)^2 = m^2 - 8m + 16$

$4 \times 4 = 16$

$4 \times \left(-\frac{m^2}{4}\right) = -m^2$

So:

$(m - 4)^2 - 16 + m^2 = 0$

Substituting $(m - 4)^2$:

$m^2 - 8m + 16 - 16 + m^2 = 0$

$2m^2 - 8m = 0$

$m(m - 4) = 0$

This gives two possible values: $m = 0$ and $m = 4$.

3. Find the equations of the tangents:

If $m = 0$:

$y = 0$

If $m = 4$:

$y = 4x - \frac{(4)^2}{4} = 4x - 4$

So the common tangents are $y = 0$ and $y = 4x - 4$. $y = 4x - 4$ can also be written as $y = 4(x - 1)$.