<div style='float: left; width:100%;'> <h3 id="successconic-section-lecture-5success"><Success>Conic section lecture-5</Success></h3> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Standard forms of parabola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>$y^2=4 a x ;$</p> <p>$y^2=-4 a x ;$</p> <p>$ x^2=4 a y ; $</p> <p>$x^2=-4 a y$</p> </div> <div style='float: right; width:40%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b534ee79-4e53-446b-a812-7af378de7f00/public"></p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/267eb0a2-27f6-4afd-4a50-2b49ce469f00/public"></p> </div>

### <Success>Equation of parabola</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Consider the general quadratic $y=a x^2+b x+c$, where $a \neq 0$.</p> <p>$y =a\left[x^2+\frac{b}{a} x+\frac{c}{a}\right] $</p> <p>$=a\left[\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4 a^2}+\frac{c}{a}\right] $</p> <p>$=a\left(x+\frac{b}{2 a}\right)^2-\frac{b^2}{4 a}+c $</p> <p>$y =a\left(x+\frac{b}{2 a}\right)^2-\left(\frac{b^2-4 a c}{4 a}\right)$</p> <p>i.e., $\quad y+\frac{b^2-4 a c}{4 a}=a\left(x+\frac{b}{2 a}\right)^2$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Equation of parabola</Success> <div style='float: left;text-align:left; width:60%;font-size:24px;'> <p>This is the form $y-k=a(x-h)^2$</p> <p>where, $$k=\frac{4ac-b^2}{4a} \quad and \quad h=\frac{-b}{2a}\quad a > 0$$</p> <p>$\alpha-k=|l-k|=k-l $</p> <p>$\Rightarrow \alpha+l=2 k$</p> <p>$P \equiv(x, \alpha) $</p> <p>$y-k=a(x-h)^2 $ <br> $\alpha-k=a(x-h)^2$ <br> $$\Rightarrow(x-h)^{2}=\frac{\alpha-k}{a} \Rightarrow x=h \pm \sqrt{\frac{\alpha-k}{a}}.$$</p> </div> <div style='float: right; width:40%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/4b3af80e-7381-4eb7-a04c-e2e0ec44b900/public"></p> </div>

### <Success>Equation of parabola</Success> <div style='float: left;text-align:left; width:100%;font-size:24px;'> <p>$P \equiv\left(h+\sqrt{\frac{\alpha-h}{a}}, \alpha\right) $</p> <p>$P F=\sqrt{\frac{\alpha-h}{a}} $</p> <p>$P l=\alpha-l$</p> <p>$ P F=P l \Rightarrow \frac{\alpha-k}{a}=(\alpha-l)^2-(i) $</p> <p>(i) $\rightarrow l=2 k-\alpha$</p> <p>Putting in (ii) $\Rightarrow \frac{\alpha-k}{a}=[\alpha-(3 k-\alpha)]^2$</p> <p>$ \Rightarrow \alpha-k=a \cdot 4(\alpha-k)^2 $</p> <p>$ \Rightarrow \alpha-k=\frac{1}{4 a} \quad[\because \alpha-k \neq 0] $</p> <p>$ \Rightarrow \alpha=k+\frac{1}{4 a}$</p> <p>$ \therefore l=2 k-\left(k+\frac{1}{4 a}\right)=k-\frac{1}{4 a}$</p> </div> <div style='float: right; width:1%;'> <!----> </div>

#### <Success>Vertex, focus and directrix of parabola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>So, for the parabola $y-k=a(x-h)^2$,</p> <p>the vertex is at $(h,k)$</p> <p>Focus is at $(h,k)$ = $(h,k-\frac{1}{4a}$</p> <p>and directrix is the line $y=l$</p> <p>i.e., $y= (k-\frac{1}{4a})$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a2cf92fd-a17b-474e-3bed-49d507d4df00/public"></p> </div>

#### <Success>Ellipse in terms of focus and directrix</Success> <div style='float: left;text-align:left; width:50%;font-size:27px;'> <p>Ellipse: $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$</p> <p>$ a>b, c^2=a^2-b^2 $</p> <p>Consider the line:</p> <p>$ x=\frac{a^2}{c}>a \quad\left(\because \frac{a}{c}>1\right) $</p> <p>Distance of $P(x, y)$ on the ellipse</p> <p>to $F_1 \equiv(c, 0) $</p> <p>$P F_1 =\sqrt{(x-c)^2+y^2}$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/2154df7d-303c-4431-77bc-6f376f133a00/public"></p> </div>

#### <Success>Ellipse in terms of focus and directrix</Success> <div style='float: left;text-align:left; width:100%;font-size:28px;'> <p>$\Rightarrow P F_1^2 =(x-c)^2+y^2 $</p> <p>$ =x^2-2 c x+c^2+y^2 \quad\left[y^2=b^2\left(1-\frac{x^2}{a^2}\right)\right] $</p> <p>$ =x^2-2 c x+c^2+b^2\left(1-\frac{x^2}{a^2}\right) $</p> <p>$ =\left(1-\frac{b^2}{a^2}\right) x^2-2 c x+c^2+b^2 $</p> <p>$ =\frac{a^2-b^2}{a^2} x^2-2 c x+a^2 \quad\left(\because c^2+b^2=a^2\right) $</p> <p>$ =\frac{c^2}{a^2} x^2-2 c x+a^2 $</p> <p>$ =\left(\frac{c}{a} x-a\right)^2=\frac{c^2}{a^2}\left(x-\frac{a^2}{c}\right)^2$</p> </div> <div style='float: right; width:1%;'> <!----> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Directrices of the ellipse</Success> <div style='float: left;text-align:left; width:50%;font-size:27px;'> <p>$ \therefore \quad P F_1^2=\left(\frac{c}{a}\right)^2 P l^2 $</p> <p>$ \Rightarrow P F_1=\frac{c}{a} P l_1=e P l_1 $</p> <p>$ \therefore \quad \frac{P F_1}{P l}=e$</p> <p>Similarly for te line $x=-\frac{-a^2}{c}$, the ratio $=e$</p> <p>Then lines $x= \pm \frac{a^2}{c}$ are called the directrices of the ellipse</p> <p>$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $a>b$.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/206e3fdc-b825-45bd-4104-6e6c5876e000/public"></p> </div>

### <Success>Eccentricity</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>we can define ellipse (hyperbola) using directory Lets take a conic section having a vetex at $(0,0)$, focus on the x-axis and directrix a line parallel to to y-axis.</p> <p>Let’s find the locus of points $P$ such that $\frac{P F}{P l}=e \leftarrow$ constant</p> <p>Since $V=(0,0)$ lies on the curve,</p> <p>$ \frac{V f}{V R}=e^{\prime} \Rightarrow V e=\frac{f}{e} $</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/370a111a-566f-4301-371d-ee7cbe8ffb00/public"></p> </div>

### <Success>Eccentricity</Success> <div style='float: left;text-align:left; width:50%;font-size:27px;'> <p>$\therefore$ the line l is $x=\frac{-f}{e}$</p> <p>$P F^2=e^2 P l^2$</p> <p>$(x-f)^2+y^2=e^2\left(x+\frac{f}{e}\right)^2 $</p> <p>$ \Rightarrow x^2-2 f x+f^2+y^2$</p> <p>$$=(e x+f)^2 =e^2 x^2+2 e f x+f^2 $$</p> <p>$$\Rightarrow\left(1-e^2\right) x^2-2 f(1+e) x+y^2=0$$</p> <p>Putting $p=f(1+e)$, we get</p> <p>$ \left(1-e^2\right) x^2-2 p x+y^2=0 $</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7c38ef0e-a0c5-48a7-810a-e91e6ccbfb00/public"></p> </div>

### <Success>Equation of parabola and ellipse</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>If $e=1$, we get $y^2=2 p x$, which is a parabola.</p> <p>If $e<1, a^2 x^2-2 p x+y^2=0$</p> <p>$ \Rightarrow \quad x^2-\frac{2 p}{a^2} x+\frac{y^2}{a^2}=0 $</p> <p>$ \Rightarrow\left(x-\frac{p}{a^2}\right)^2-\frac{p^2}{a^4}+\frac{y^2}{a^2}=0 $</p> <p>$ \Rightarrow\left(x-\frac{p}{a^2}\right)^2+\frac{y^2}{a^2}=\frac{p^2}{a^4}$</p> <p>which is equation of the ellipse</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/6af10108-a1a8-4184-def7-7410ac49ae00/public"></p> </div>

### <Success>Equation of circle and hyperbola</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Special case : $e=0$ gives</p> <p>$ x^2-2 p x+y^2=0 $</p> <p>which is equation of a circle.</p> <p>If $e>1$,</p> <p>$ {(e^2-1)} x^2+2 p x-y^2=0 $</p> <p>which is equation of a hyperbola.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Equation of a hyperbola</Success> <div style='float: left;text-align:left; width:70%;font-size:28px;'> <p>$$\text{Consider the hyperbola:} \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$</p> <p>$ \text { Line : } x=\frac{a^2}{c} $</p> <p>$c^2=a^2+b^2>a^2 $</p> <p>$\frac{a}{c}<1 $</p> <p>$\therefore \frac{a^2}{c}<a $</p> <p>$ P F_1^2=(x-c)^2+y^2=x^2-2 c x+c^2+y^2 $</p> <p>$y^2 =b^2\left(\frac{x^2}{a^2}-1\right)=\frac{b^2}{a^2} x^2-b^2$</p> </div> <div style='float: right; width:30%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/c5f7eeea-1c5f-4b65-7c3c-d19eca2f7d00/public"></p> </div>

### <Success>Directrix of the hyperbola</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\therefore P F_1^2 =x^2-2 c x+c^2+\frac{b^2}{a^2} x^2-b^2 $</p> <p>$=\left(1+\frac{b^2}{a^2}\right) x^2-2 c x+\left(c^2-b^2\right) $</p> <p>$=\frac{c^2}{a^2} x^2-2 c x+a^2 $</p> <p>$=\left(\frac{c}{a} x-a\right)^2=\left(\frac{c}{a}\right)^2\left(x-\frac{a^2}{c}\right)^2 $</p> <p>$=e^2 P l^2 $</p> <p>$\therefore \quad \frac{P F_1}{P l} =e>1$</p> <p>So, th line $x= \pm \frac{a^2}{c}$ are the directrix of the hyperbola</p> <p>$\frac{x^2}{a^2}-\frac{y^2}{x^2}=1$.</p> </div> <div style='float: right; width:1%;'> <!----> </div>

<div> <h2><Success>Thank you</Success></h2> </div>