### <Success>Conic sections lecture-3 </Success> <div style='float: left; width:100%;'> </div> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Hyperbola</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>A hyperbola in the set of all points in a plane, the difference of whose distances from two fixed points in the plane is a constant (less than to distance between th two fixed points)</p> <p>The two fixed points are called the foci of the hyperbola.</p> <p>The mid point of the line segment joining the foci is called the centre.</p> <div style='float: right; width:1%;'> <!----> <!----> </div>

### <Success>Axis of the hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <ul> <li>The line through the foci is called, the transverse axis of the hyperbola</li> </ul> <p>The line perpendicular to the transverse axis and passing through the centre is calls the conjungate axis of the hyperbola.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/85762534-f2e7-4f9c-6b94-c50b5b120000/public"></p> </div>

### <Success>Hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:25px;'> <p>Let the distance between the two foci,</p> <p>$ F_1 F_2=2 c>0 $</p> <p>let centre be at the origin $(0,0)$ and the foci lie on the x axis.</p> <p>Then $F_1 \equiv(-c, 0)$</p> <p>$ F_2 \equiv(c, 0) $</p> <p>If $P$ is end point on the hyperbola, then</p> <p>$\left|P F_1-P F_2\right|=$ constant</p> <p>$ =2 a \text { (say) } $</p> <p>we have $a<c$</p> </div> <div style='float: right; width:40%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/c3f94a3c-9c2a-4388-5041-dd31caddf400/public"></p>

### <Success>Hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>Let $A=(x, 0)$ be a point on the hyperbola</p> <p>If $0<x<c$,</p> <p>$ A F_1-A F_2=2 a $</p> <p>but</p> <p>$A F_1 =c+x $</p> <p>$A F_2 =c-x $</p> <p>$A F_1-A F_2 =(c+x)-(c-x)=2 x$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/16c8cf6a-79fe-4fa6-733e-cde5bbbefc00/public"></p> <p>======</p> <h3 id="successhyperbolasuccess"><Success>Hyperbola</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>So, $2 x=2 a \Rightarrow x=a$.</p> <p>So, $(a, 0)$ lies on the hyperbola.</p> <p>similarly, $(-a, 0)$ also lies on the hyperbola</p> <p>Also, if $|x| \geqslant c$, than $(x, 0)$ does not lie on the hyperbola.</p> </div>

### <Success>Hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>Because if $x \geqslant c$ on $x \leqslant-c$, then the diff. of distances from</p> <p>$F_1 to F_2$ is $F_1 F_2=2 c \neq 2 a$</p> <p>$A F_1-A F_2=F_1 F_2$</p> <p>$B F_2-8 F_1=F_1 F_2$.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d04584c9-a474-4749-23ce-eec32e79bc00/public"></p> </div>

### <Success> Vertices of the hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>So, there are exactly two points on the transverse axis which lie on the hyperbola</p> <p>The coordinates of these two points are $ (-a, 0) \text { and }(a, 0) \text {. } $</p> <p>There two points are called the vertices of the hyperbola</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fa03e06d-4ec5-4ab3-8f41-ec0236bf0a00/public"></p> </div>

### <Success>Standard forms of hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <ol> <li> <p>Foci on the x-axis.</p> </li> <li> <p>Foci on the y-axis.</p> </li> </ol> <p>(1) Let $P=(x,y)$ in any point on the hyperbola with $x>0$</p> <p>then $PF_1 - PF_2 = 2a$</p> <p>$PF_1 = \sqrt{(x+c)^2+y^2}$</p> <p>$PF_2 = \sqrt{(x-c)^2+y^2}$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/04ea5b87-3c79-497f-a4f4-9a026d040000/public"></p> </div>

### <Success>Standard forms of hyperbola</Success> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>So, $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2 a$</p> <p>$\Rightarrow \quad(x+c)^2+y^2=\left[2 a+\sqrt{(x-c)^2+y^2}\right]^2 $</p> <p>$\Rightarrow \quad(x+c)^2+y^2=4 a^2+(x-c)^2+y^2+4 a \sqrt{(x-c)^2+y^2} $</p> <p>$\Rightarrow 4a \sqrt{(x-c)^2+y^2}=(x+c)^2-(x-c)^2-4 a^2 $ $=4 c x-4 a^2 $</p> <p>$\Rightarrow a^2\left[(x-c)^2+y^2\right]=\left(c x-a^2\right)^2 $</p> <p>$\Rightarrow a^2 x^2+a^2 c^2-2 a^2 c x+a^2 y^2=c^2x^2+a^4-2 a^2 cx $</p> <p>$\Rightarrow\left(c^2-a^2\right) x^2-a^2 y^2=a^2 c^2-a^4=a^2\left(c^2-a^2\right) $</p> <p>$\text { Let’s put } c^2-a^2=b^2$</p> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Standard forms of hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>then $b^2 x^2-a^2 y^2=a^2 b^2$</p> <p>Dividing by $a^2 b^2$,</p> <p>we get $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1 \leftarrow {\text { symmetric about x-axis and y-axis }}$</p> <p>$ \frac{x^2}{a^2}=1+\frac{y^2}{b^2} \geqslant 1 \Rightarrow x^2 \geqslant a^2 $</p> <p>$\Rightarrow x \geqslant a \quad(for x>0)$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1e0c622f-0f36-4478-b9d1-2824c3c61500/public"></p> </div>

### <Success>Standard forms of hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>Similarly if $P \equiv(x, y)$ with $x<0$ on th Hyperbola, the</p> <p>$ P F_2-P F_1=2 a $</p> <p>Proceeding like in previous case, we get the same equaion.</p> <p>$ \frac{x^2}{a^2}-\frac{y^2}{b^2}=1 $</p> <p>where $b^2=c^2-a^2$</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/783cb4de-aa3e-4d32-0b2e-a8a155e4b100/public"></p> </div>

### <Success>Standard forms of hyperbola</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>So the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ looks like</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/799494a7-0006-4dfb-4db7-02142bfcf700/public"></p> </div>

### <Success>Remark</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <ul> <li>the hyperbola is symmetric about the transverse axis so will as the conjugate axis.</li> </ul> <p>Focii on the y-axis.</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/28437031-16c4-4f4f-289f-f2a1ffddf800/public"></p> </div>

### <Success>Equation of hyperbola</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>The equation of the hyperbola with focii on the y-axis is given by</p> <p>$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$</p> <p>This intersect the y-axis at $(0,\pm a)$</p> <p>Also, $|y| \geqslant a$ se. $y \geqslant a$ or $y \leqslant-a$</p> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/cfdf1dac-4edf-4969-bd6b-0adba6003d00/public"></p> </div>

### <Success>Latus rectum</Success> <div style='float: left;text-align:left; width:50%;font-size:28px;'> <p>Line segment passing through on focus, perpendicular to the transverse axis, and having end points on the hyperbola</p> <p>Let $A \equiv(c,-\beta) \quad and B \equiv(c, \beta)$</p> <p>Then length of the latus rectum, $l=2 \beta$</p> <p>since $(c, \beta)$ lies on th hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$,</p> <p>we get</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8b70dc04-160e-416e-7832-36f140063100/public"></p> </div>

### <Success>Latus rectum</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\frac{c^2}{a^2}-\frac{k^2}{b^2}=1 \Rightarrow \frac{p^2}{b^2}=\frac{c^2}{a^2}-1$</p> <p>$=\frac{c^2-a^2}{a^2}=\frac{b^2}{a^2} $</p> <p>$ \Rightarrow \beta^2=\frac{b^4}{a^2} \Rightarrow \beta=\frac{b^2}{a}$</p> <p>So, $l=\frac{2 b^2}{a}$</p> <p>We define the eccentricity of the hyperbola as</p> <p>$ e = \frac{c}{a} >1$</p> <div style='float: right; width:1%;'> <!----> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>Find the foci, vertices, eccentricity and length of latus rectum for</p> <p>(i) $\frac{x^2}{16}-\frac{y^2}{9}=1$</p> <p>(ii) $9 y^2-4 x^2=36$</p> <p>(iii) $\quad \frac{x^2}{16}-\frac{y^2}{9}=1 \Rightarrow a=4, b=3$</p> <p><strong>Solution :</strong></p> <p>Foci lies on $x$-axis.</p> <p>$ c^2-a^2=b^2 \Rightarrow c^2=a^2+b^2=4^2+3^2=25 $</p> <p>$\therefore$ Foci at $( \pm 5,0)$</p> <p>Vertices at $( \pm a, 0)=( \pm 4,0)$</p> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/1ece23a1-70bd-4fdb-609b-1e2168e67c00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:23px;'> <p>$ \begin{aligned} & \therefore c=\frac{c}{a}=\frac{5}{4} \ & =l=\frac{2 b^2}{a}=2 \times \frac{9}{4}=\frac{9}{2} . \end{aligned} $</p> <p>(ii)</p> <p>$9 y^2-4 x^2 =36 $</p> <p>$\quad \frac{y^2}{4}-\frac{x^2}{9} =1 $</p> <p>$ \text { This is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$ } $ <br> $\text { with}$ $a =2, b=3$</p> <p>Focii lie on th $y$-axis $(0, \pm c)$</p> <p>$c^2=a^2+b^2=2^2+3^2=13 $ <br> $c=\sqrt{17}$</p> <div style='float: right; width:1%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/dc3cd643-ccbc-466c-df67-75c44bd29c00/public"></p> </div>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\therefore \text { Foci } \equiv(0, \pm \sqrt{13})$</p> <p>$\text { Vertices } \equiv(0, \pm a)=(0, \pm 2)$</p> <p>$e=\frac{c}{a}=\frac{\sqrt{12}}{2}$</p> <p>$l=\frac{2 b^2}{a}=\frac{2 \times x^2}{2}=9 $</p> <div style='float: right; width:1%;'> <!----> </div>

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