<div style='float: left; width:100%;'> <h3 id="successconic-section-lecture---7success"><Success>Conic section lecture - 7</Success></h3> </div> <div style='float: right; width:1%;'> <!----> </div>

<h3 id="successparametric-form-of-the-parabolasuccess"><Success>Parametric form of the parabola</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p><strong>Parametric form of the parabola $y^2=4ax$</strong></p> <p>If we put $y=t \in \R$,</p> <p>then, $x = \frac{y^2}{4a} = \frac{t^2}{4a}$</p> <p>So, any general point on the parabola can be written as $x=\frac{t^2}{4a}$ and $y=t$</p> <p>Instead of $y=t$ if we use $y=2at$</p> <p>then $x=\frac{(2at)^2}{4a} = \frac{4a^2t^2}{4a} = at^2$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/aeb932eb-8123-4bf4-931e-f45fd275ce00/public"></p> </div>

<h3 id="successparametric-form-of-the-parabolasuccess-1"><Success>Parametric form of the parabola</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>We will use $x=at^2,y=2at$ as the parametric form of the parabola $y^2=4ax$.</p> <p>i.e any general point of the parabola $y^2=4ax$ can be written as $(at^2, 2at)$.</p> <p>As t various between $-\infty$ and $+\infty$, we get all points on the parabola.</p> <p>Also, there is a unique point $(at^2, 2at)$ on the parabola for any given $t \in \R$.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/aa51c159-92b3-4c71-a24b-61c911bc8600/public"></p> </div>

<h3 id="successequation-of-tangent-and-normal-success"><Success>Equation of tangent and normal </Success></h3> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p><strong>Equation of tangent & normal in parametric form</strong>:</p> <p>Recall : equation of tangent at $(x_1, y_1)$ is given by</p> <p>$y y_1=2 a(x+x_1)$</p> <p>Equation of normal at $(x_1, y_1)$ is given by</p> <p>$y-y_1=\frac{-y_1}{2 a} (x-x_1)$</p> <p>Using $(x_1, y_1)=(a t^2, 2 a t)$,</p> <p>we get Equation of tangent:</p> <p>$y.2 a t=2 a(x+a t^2)$</p> <p>that is $t y=x+a t^2$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/3aad666f-778f-4959-6d00-c31507f92e00/public"></p> </div>

<h3 id="successequation-of-tangent-and-normalsuccess"><Success>Equation of tangent and normal</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$y$-2at $=-\frac{2 a t}{3 a}(x-a t^2)$</p> <p>ie. $y-2 a t=-t x+a t^3$</p> <p>ie. $y+t x=2 a t+a t^3$</p> <p>slope $=-t$.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<h3 id="successequation-of-the-chord-success"><Success>Equation of the chord </Success></h3> <div style='float: left;text-align:left; width:50%;font-size:25px;'> <p>Equation of the chord joining $P \equiv (at_1^2, 2at_1)$</p> <p>and $Q \equiv (at_2^2, 2at_2)$</p> <p>$y-2at_1 = \frac{2at_1-2at_2}{at_1^2-at_2^2} (x-at_1^2)$</p> <p>$\Leftrightarrow y-2at_1 = \frac{2}{t_1+t_2} (x-at_1^2)$</p> <p>$\Leftrightarrow (t_1+t_2)y - 2at_1 (t_1+t_2) = 2(x-at_1^2)$</p> <p>$\Leftrightarrow y(t_1+t_2) = 2at_1^2 + 2a t_1 t_2 + 2x - 2at_1^2$</p> <p>$\Leftrightarrow y(t_1+t_2) = 2(x+at_1t_2)$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b72ef545-9d52-46a1-2e75-fc286710da00/public"></p> </div>

<h4 id="successpoint-of-intersection-of-tangents-on-the-parabola-success"><Success>Point of intersection of tangents on the parabola </Success></h4> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>Point of intersection of tangents at two different point on the parabola $y^2=4 a x$.</p> <p>Let $P \equiv(a t_1^2, 2 a t_1)$ & $Q \equiv(a t_2^2, 2 a t_2)$</p> <p>Tangent at $P: \quad t, y=x+a t_1^2$</p> <p>Target at Q : $t_2 y=x+a t_2^2$</p> <p>$(i)-(ii) \Rightarrow (t_1-t_2) y=a(t_1^2-t_2^2)$</p> <p>$\Rightarrow y=a(t_1+t_2) $</p> <p>$\therefore x =t_1 y-a t_1^2=a t_1(t_1+t_2)-a t_1^2 = a t_1 t_2$</p> <p>Point of intersection is $(a t_1 t_2, a(t_1+t_2))$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0985f7bc-235a-498f-b461-c98f383a9400/public"></p> </div>

<h3 id="successtheoremsuccess"><Success>Theorem</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>If the tangents at point $P$ and $Q$ on a parabola meet at point $T$, prove that</p> <p>(i) $T P$ and $T Q$ subtend equal angle at the focus S .</p> <p>(ii) $S T^2=S P \cdot S Q$</p> <p>(iii) $\triangle S P T$ is similar to $\triangle S T Q$.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/03406fef-15d3-42bd-4220-5168808c2200/public"></p> </div>

<h3 id="successproofsuccess"><Success>Proof</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>let $P \equiv(a t_1^2, 2 a t_1)$ & $Q \equiv(a t_2^2, 2 a t_2)$ By the previous formula for the point of intersection of tangent lives at $P$ & $Q$, we have</p> <p>$T \equiv(a t_1 t_2, a(t_1+t_2))$</p> <p>Equation of the line $S P$</p> <p>$ y-0=\frac{2 a t_1-0}{a t_1^2-a}(x-a) $</p> <p>$[\therefore S \equiv (a,0) and P \equiv (at_1^2,2at_1)]$</p> <p>$\Leftrightarrow y=\frac{2 t_1^2}{t_1^2-1}(x-a) $</p> <p>$\Leftrightarrow (t_1^2-1) y-2 t_1 x+2 a t_1=0$</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/74117a22-00ab-4dbf-4790-3e864252a900/public"></p> </div>

<h3 id="successproofsuccess-1"><Success>Proof</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Now let M be the foot of perpendicular from T to the Joining S & P.</p> <p>Recall: The perpendicular distance of a point $(x_1,y_1)$ to the line $ax+bx+c=0$ is given by</p> <p>$\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}$</p> <p>$\therefore TM = \frac{|(t_1^2-1) a(t_1+t_2)-2t_1 a t_1t_2+2at_1|}{\sqrt{(t_1^2-1)^2 + (2t_1)^2}}$</p> <p>$= \frac{a|(t_1^2-1)(t_1+t_2)-2t_1^2t_2+2t_1|}{\sqrt{(t_1^2+1)^2}}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<h3 id="successproofsuccess-2"><Success>Proof</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\therefore T M =\frac{a|t_1^3-t_1+t_1^2 t_2-t_2-2 t_1^2 t_2+2 t_1|}{t_1^2+1} $</p> <p>$ =\frac{a|t_1^3-t_1^2 t_2+(t_1-t_2)|}{t_1^2+1} $</p> <p>$ =\frac{a|(t_1-t_2)(t_1^2+1)|}{(t_1^2+1)}=a|t_1-t_2|$</p> <p>Since this is symmetric w.r.t. $t_1$ & $t_2$,</p> <p>$T M’=T M=a|t_1-t_2|$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<h3 id="successproofsuccess-3"><Success>Proof</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>$TM=TM'$</p> <p>$\triangle T M S$ and $\triangle T M’ S$ are congruent</p> <p>$\Rightarrow \angle T S M=\angle T S M’$ ie. $\angle T S P=\angle T S Q$.</p> <p>This proves (i).</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/251220c2-79d7-4259-6fa3-003abeeed800/public"></p> </div>

<h3 id="successto-provesuccess"><Success>To prove</Success></h3> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <p>$ST^2 = (SP)(SQ)$</p> <p>$SP^2 = (at_1^2-a)^2 + (2at_1-0)^2 = a^2(t_1^2-1)^2+4a^2t_1^2 = a^2 (t_1^2+1)^2$</p> <p>$SP = a(t_1^2+1)$</p> <p>Similarly $SQ = a(t_2^2+1)$</p> </div> ====== <h3 id="successto-provesuccess-1"><Success>To prove</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$ST = (at_1t_2-a)^2 + (a(t_1+t_2)-0)^2$</p> <p>$= a^2[(t_1t_2-1)^2+(t_1+t_2)^2]$</p> <p>$= a^2 [t_1^2t_2^2+1-2t_1t_2+t_1^2+t_2^2+2t_1t_2]$</p> <p>$= a^2 (t_1^2 +1)(t_2^2+1)$</p> <p>$= (SP)(SQ)$, which proves (ii)</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a5f34fdf-e0d1-4973-37e5-25f344b9a600/public"></p> <!----> </div>

<h3 id="successto-provesuccess-2"><Success>To prove</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>Next, to show $\Delta SPT \sim \Delta STQ$.</p> <p>We have from (i) & (ii)</p> <p>$\angle PST = \angle QST$</p> <p>and $\frac{ST}{SP}=\frac{SQ}{ST}$</p> <p>$\therefore \Delta SPT$ is similar to $\Delta STQ$.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/0b01fee0-ff96-434e-94fd-551bab73fb00/public"></p> </div>

<h3 id="successexamplesuccess"><Success>Example</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>prove that the area of the triangle formed by three points on a parabola is twice the area of the triangle formed by the tangents at these points.</p> <p><strong>Solution</strong>. We can assume that the equation of the parabola is $y^2=4 a x$ (by choosing the vertex to be at $(0,0)$ and to axis as the $x$-axis.)</p> <p>let $P \equiv (a t_1^2, 2 a t_1)$</p> <p>$Q \equiv (a t_2^2, 2 a t_2)$</p> <p>$R \equiv (a t_3^2 ,2 a t_3)$</p> <p>be any 3 points on the parabola.</p> </div> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/3ef69ac4-a6ec-453b-e827-5699477aee00/public"></p> </div>

<h3 id="successsolutionsuccess"><Success>Solution</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:25px;'> <p><strong>Recall</strong>: Area of the triangle formed by $(x_1,y_1),(x_2,y_2)$ & $(x_3,y_3)$</p> <p>Area of $\Delta PQR$ = area of trap. PABR + area of trap. RBCQ - area of trap. PACQ</p> <p>$= \frac{1}{2} (y_1+y_2) (x_3-x_1) + \frac{1}{2} (y_2+y_3) (x_2-x_3) - \frac{1}{2} (y_1+y_2) (x_2-x_1)$</p> <p>$\Delta = \frac{1}{2} [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$</p> <p>=$\frac{1}{2}\left|\begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\x_2 & y_2 & 1\end{array}\right|$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/eb382bc6-dbaf-4e09-0f42-76990637b700/public"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Then area of $\triangle P Q R$</p> <p>$ =\frac{1}{2} [a t_1^2(2 a t_2-2 a t_3)+a t_3^2 (2 a t_3-2 a t_1) + a t_3^2(2 a t_1-2 a t_2)] $</p> <p>$ =a^2[t_1^2(t_2-t_3)+t_2^2(t_3-t_1)+t_3^2(t_1-t_2$</p> <p>$ =-a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)$ (eqn-1)</p> <p>Points of intersection of tangent are</p> <p>$(a t_1 t_3, a(t_1+t_2))$, $(a t_2 t_3, a(t_2+t_3))$, $(a t_3 t_1, a(t_3+t_1))$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Area of $\Delta$ formed by these tangent</p> <p>$ = \frac{1}{2} [at_1t_2 (at_2-at_1) + at_2t_3 (at_3-at_2) + at_3t_1 (at_1-at_3)]$</p> <p>$ = \frac{1}{2} a^2 (t_1-t_2)(t_2-t_3)(t_1-t_3)$</p> <p>$ = \frac{1}{2} \text{(area of PQR)}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<div> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> </div>