Parabola

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 $y = 4(x - 1)$.