Class 11: Parabola – Exercise 25.1

Note:

	$\displaystyle y^2 = 4ax$	$\displaystyle y^2 = 4ax$	$\displaystyle x^2 = 4ay$	$\displaystyle x^2 = - 4ay$
Coordinates of Vertex	$\displaystyle (0,0)$	$\displaystyle (0,0)$	$\displaystyle (0,0)$	$\displaystyle (0,0)$
Coordinates of focus	$\displaystyle (a, 0)$	$\displaystyle (a, 0)$	$\displaystyle (0, a)$	$\displaystyle (0, -a)$
Equation of the directrix	$\displaystyle x = - a$	$\displaystyle x=a$	$\displaystyle y=-a$	$\displaystyle y=a$
Equation of the axis	$\displaystyle y=0$	$\displaystyle y=0$	$\displaystyle x=0$	$\displaystyle x=0$
Length of Latus rectum	$\displaystyle 4a$	$\displaystyle 4a$	$\displaystyle 4a$	$\displaystyle 4a$
Focal distance of a point $\displaystyle P(x, y)$	$\displaystyle a+x$	$\displaystyle a-x$	$\displaystyle a+y$	$\displaystyle a-y$

$\displaystyle \textbf{Question 1: }\text{Find the equation of the parabola whose:}$
$\displaystyle \text{(i) focus is }(3,0)\text{ and the directrix is }3x+4y=1;$
$\displaystyle \text{(ii) focus is }(1,1)\text{ and the directrix is }x+y+1=0;$
$\displaystyle \text{(iii) focus is }(0,0)\text{ and the directrix is }2x-y-1=0;$
$\displaystyle \text{(iv) focus is }(2,3)\text{ and the directrix is }x-4y+3=0.$
$\displaystyle \text{Answer:}$

$\displaystyle \text{(i) Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{The focus is }S(3,0)\text{ and the directrix is }3x+4y-1=0.$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-3)^2+y^2=\left(\frac{3x+4y-1}{\sqrt{3^2+4^2}}\right)^2$
$\displaystyle (x-3)^2+y^2=\frac{(3x+4y-1)^2}{25}$
$\displaystyle 25\left[(x-3)^2+y^2\right]=(3x+4y-1)^2$
$\displaystyle 25x^2-150x+225+25y^2$
$\displaystyle =9x^2+16y^2+1+24xy-6x-8y$
$\displaystyle 16x^2+9y^2-24xy-144x+8y+224=0$
$\displaystyle \therefore \text{The required equation is }16x^2+9y^2-24xy-144x+8y+224=0.$

$\displaystyle \text{(ii ) Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{The focus is }S(1,1)\text{ and the directrix is }x+y+1=0.$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix.}$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-1)^2+(y-1)^2=\left(\frac{x+y+1}{\sqrt{1^2+1^2}}\right)^2$
$\displaystyle (x-1)^2+(y-1)^2=\frac{(x+y+1)^2}{2}$
$\displaystyle 2\left[(x-1)^2+(y-1)^2\right]=(x+y+1)^2$
$\displaystyle 2x^2+2y^2-4x-4y+4$
$\displaystyle =x^2+y^2+1+2xy+2x+2y$
$\displaystyle x^2+y^2-2xy-6x-6y+3=0$
$\displaystyle \therefore \text{The required equation of the parabola is}$
$\displaystyle x^2+y^2-2xy-6x-6y+3=0.$
$\displaystyle \\$

$\displaystyle \text{(iii) Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{The focus is }S(0,0)\text{ and the directrix is }2x-y-1=0.$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle x^2+y^2=\left(\frac{2x-y-1}{\sqrt{2^2+(-1)^2}}\right)^2$
$\displaystyle x^2+y^2=\frac{(2x-y-1)^2}{5}$
$\displaystyle 5x^2+5y^2=(2x-y-1)^2$
$\displaystyle 5x^2+5y^2=4x^2+y^2+1-4xy-4x+2y$
$\displaystyle x^2+4y^2+4xy+4x-2y-1=0$
$\displaystyle \therefore \text{The required equation is }x^2+4y^2+4xy+4x-2y-1=0.$

$\displaystyle \text{(iv) Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{The focus is }S(2,3)\text{ and the directrix is }x-4y+3=0.$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-2)^2+(y-3)^2=\left(\frac{x-4y+3}{\sqrt{1^2+(-4)^2}}\right)^2$
$\displaystyle (x-2)^2+(y-3)^2=\frac{(x-4y+3)^2}{17}$
$\displaystyle 17\left[(x-2)^2+(y-3)^2\right]=(x-4y+3)^2$
$\displaystyle 17x^2+17y^2-68x-102y+221$
$\displaystyle =x^2+16y^2+9-8xy+6x-24y$
$\displaystyle 16x^2+y^2+8xy-74x-78y+212=0$
$\displaystyle \therefore \text{The required equation is }16x^2+y^2+8xy-74x-78y+212=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Find the equation of the parabola whose focus is the point }(2,3)$
$\displaystyle \text{and whose directrix is the line }x-4y+3=0.\text{ Also, find the length of its}$
$\displaystyle \text{latus rectum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{The focus is }S(2,3)\text{ and the directrix is }x-4y+3=0.$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix.}$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-2)^2+(y-3)^2=\left(\frac{x-4y+3}{\sqrt{1^2+(-4)^2}}\right)^2$
$\displaystyle (x-2)^2+(y-3)^2=\frac{(x-4y+3)^2}{17}$
$\displaystyle 17\left[(x-2)^2+(y-3)^2\right]=(x-4y+3)^2$
$\displaystyle 17x^2+17y^2-68x-102y+221$
$\displaystyle =x^2+16y^2+9-8xy+6x-24y$
$\displaystyle 16x^2+y^2+8xy-74x-78y+212=0$
$\displaystyle \therefore \text{The required equation of the parabola is}$
$\displaystyle 16x^2+y^2+8xy-74x-78y+212=0.$
$\displaystyle \text{The perpendicular distance of the focus from the directrix is }2a.$
$\displaystyle 2a=\left|\frac{(1)(2)+(-4)(3)+3}{\sqrt{1^2+(-4)^2}}\right|$
$\displaystyle =\left|\frac{-7}{\sqrt{17}}\right|=\frac{7}{\sqrt{17}}$
$\displaystyle \text{Length of the latus rectum}=4a=2(2a)$
$\displaystyle =2\times\frac{7}{\sqrt{17}}=\frac{14}{\sqrt{17}}=\frac{14\sqrt{17}}{17}.$
$\displaystyle \therefore \text{The length of the latus rectum is }\frac{14}{\sqrt{17}}\text{ units.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equation of the parabola, if}$
$\displaystyle \text{(i) the focus is at }(-6,-6)\text{ and the vertex is at }(-2,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) The focus is }S(-6,-6)\text{ and the vertex is }A(-2,2).$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point where the axis meets the directrix.}$
$\displaystyle \text{The vertex }A\text{ is the midpoint of the line segment }SZ.$
$\displaystyle \therefore \frac{-6+x_1}{2}=-2$
$\displaystyle -6+x_1=-4$
$\displaystyle x_1=2$
$\displaystyle \text{Also, }\frac{-6+y_1}{2}=2$
$\displaystyle -6+y_1=4$
$\displaystyle y_1=10$
$\displaystyle \therefore Z=(2,10).$
$\displaystyle \text{The slope of the axis }AS\text{ is}$
$\displaystyle m_1=\frac{2-(-6)}{-2-(-6)}=\frac{8}{4}=2.$
$\displaystyle \text{Let }m_2\text{ be the slope of the directrix.}$
$\displaystyle \text{Since the directrix is perpendicular to the axis, }m_1m_2=-1.$
$\displaystyle \therefore m_2=-\frac{1}{2}.$
$\displaystyle \text{The directrix passes through }Z(2,10)\text{ and has slope }-\frac{1}{2}.$
$\displaystyle y-10=-\frac{1}{2}(x-2)$
$\displaystyle 2y-20=-x+2$
$\displaystyle x+2y-22=0.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix.}$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x+6)^2+(y+6)^2=\left(\frac{x+2y-22}{\sqrt{1^2+2^2}}\right)^2$
$\displaystyle (x+6)^2+(y+6)^2=\frac{(x+2y-22)^2}{5}$
$\displaystyle 5\left[(x+6)^2+(y+6)^2\right]=(x+2y-22)^2$
$\displaystyle 5x^2+5y^2+60x+60y+360$
$\displaystyle =x^2+4y^2+4xy-44x-88y+484$
$\displaystyle 4x^2+y^2-4xy+104x+148y-124=0$
$\displaystyle \therefore \text{The required equation of the parabola is}$
$\displaystyle 4x^2+y^2-4xy+104x+148y-124=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equation of the parabola, if}$
$\displaystyle \text{(ii) the focus is at }(0,-3)\text{ and the vertex is at }(0,0).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(ii) The focus is }S(0,-3)\text{ and the vertex is }A(0,0).$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point where the axis meets the directrix.}$
$\displaystyle \text{The vertex }A\text{ is the midpoint of the line segment }SZ.$
$\displaystyle \therefore \frac{0+x_1}{2}=0$
$\displaystyle x_1=0$
$\displaystyle \text{Also, }\frac{-3+y_1}{2}=0$
$\displaystyle -3+y_1=0$
$\displaystyle y_1=3$
$\displaystyle \therefore Z=(0,3).$
$\displaystyle \text{Hence, the equation of the directrix is }y=3.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix }y-3=0.$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-0)^2+\{y-(-3)\}^2=\left(\frac{y-3}{\sqrt{0^2+1^2}}\right)^2$
$\displaystyle x^2+(y+3)^2=(y-3)^2$
$\displaystyle x^2+y^2+6y+9=y^2-6y+9$
$\displaystyle x^2+12y=0$
$\displaystyle \therefore x^2=-12y.$
$\displaystyle \therefore \text{The required equation of the parabola is }x^2+12y=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equation of the parabola, if}$
$\displaystyle \text{(iii) the focus is at }(0,-3)\text{ and the vertex is at }(-1,-3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iii) The focus is }S(0,-3)\text{ and the vertex is }A(-1,-3).$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point where the axis meets the directrix.}$
$\displaystyle \text{The vertex }A\text{ is the midpoint of the line segment }SZ.$
$\displaystyle \therefore \frac{0+x_1}{2}=-1$
$\displaystyle x_1=-2$
$\displaystyle \text{Also, }\frac{-3+y_1}{2}=-3$
$\displaystyle -3+y_1=-6$
$\displaystyle y_1=-3$
$\displaystyle \therefore Z=(-2,-3).$
$\displaystyle \text{Since }S(0,-3)\text{ and }A(-1,-3)\text{ have the same ordinate,}$
$\displaystyle \text{the axis }AS\text{ is horizontal.}$
$\displaystyle \text{Therefore, the directrix is vertical and passes through }Z(-2,-3).$
$\displaystyle \therefore \text{The equation of the directrix is }x=-2,\text{ or }x+2=0.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix }x+2=0.$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-0)^2+\{y-(-3)\}^2=\left(\frac{x+2}{\sqrt{1^2+0^2}}\right)^2$
$\displaystyle x^2+(y+3)^2=(x+2)^2$
$\displaystyle x^2+y^2+6y+9=x^2+4x+4$
$\displaystyle y^2-4x+6y+5=0$
$\displaystyle \therefore (y+3)^2=4(x+1).$
$\displaystyle \therefore \text{The required equation of the parabola is }y^2-4x+6y+5=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equation of the parabola, if}$
$\displaystyle \text{(iv) the focus is at }(a,0)\text{ and the vertex is at }(a',0).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv) The focus is }S(a,0)\text{ and the vertex is }A(a',0).$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point where the axis meets the directrix.}$
$\displaystyle \text{The vertex }A\text{ is the midpoint of the line segment }SZ.$
$\displaystyle \therefore \frac{a+x_1}{2}=a'$
$\displaystyle a+x_1=2a'$
$\displaystyle x_1=2a'-a$
$\displaystyle \text{Also, }\frac{0+y_1}{2}=0$
$\displaystyle y_1=0$
$\displaystyle \therefore Z=(2a'-a,0).$
$\displaystyle \text{Since }S(a,0)\text{ and }A(a',0)\text{ lie on a horizontal line,}$
$\displaystyle \text{the axis is horizontal and the directrix is vertical.}$
$\displaystyle \therefore \text{The equation of the directrix is }x=2a'-a.$
$\displaystyle \text{Hence, the directrix is }x-(2a'-a)=0.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix.}$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle (x-a)^2+y^2=\left(\frac{x-(2a'-a)}{\sqrt{1^2+0^2}}\right)^2$
$\displaystyle (x-a)^2+y^2=\{x-(2a'-a)\}^2$
$\displaystyle x^2-2ax+a^2+y^2=x^2-2(2a'-a)x+(2a'-a)^2$
$\displaystyle y^2=4(a-a')x+4a'(a'-a)$
$\displaystyle y^2=4(a-a')(x-a')$
$\displaystyle \therefore y^2+4(a'-a)(x-a')=0.$
$\displaystyle \therefore \text{The required equation of the parabola is }y^2=4(a-a')(x-a').$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find the equation of the parabola, if}$
$\displaystyle \text{(v) the focus is at }(0,0)\text{ and the vertex is at the intersection of the lines}$
$\displaystyle x+y=1\text{ and }x-y=3.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(v) The focus is }S(0,0).$
$\displaystyle \text{The vertex is at the intersection of }x+y=1\text{ and }x-y=3.$
$\displaystyle \text{Adding the two equations,}$
$\displaystyle 2x=4$
$\displaystyle x=2.$
$\displaystyle \text{Substituting }x=2\text{ in }x+y=1,$
$\displaystyle 2+y=1$
$\displaystyle y=-1.$
$\displaystyle \therefore \text{The vertex is }A(2,-1).$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point where the axis meets the directrix.}$
$\displaystyle \text{The vertex }A\text{ is the midpoint of the line segment }SZ.$
$\displaystyle \therefore \frac{0+x_1}{2}=2$
$\displaystyle x_1=4.$
$\displaystyle \text{Also, }\frac{0+y_1}{2}=-1$
$\displaystyle y_1=-2.$
$\displaystyle \therefore Z=(4,-2).$
$\displaystyle \text{The slope of the axis }AS\text{ is}$
$\displaystyle m_1=\frac{-1-0}{2-0}=-\frac{1}{2}.$
$\displaystyle \text{Let }m_2\text{ be the slope of the directrix.}$
$\displaystyle \text{Since the directrix is perpendicular to the axis, }m_1m_2=-1.$
$\displaystyle \therefore m_2=\frac{-1}{-\frac{1}{2}}=2.$
$\displaystyle \text{The directrix passes through }Z(4,-2)\text{ and has slope }2.$
$\displaystyle y-(-2)=2(x-4)$
$\displaystyle y+2=2x-8$
$\displaystyle 2x-y-10=0.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the required parabola.}$
$\displaystyle \text{Draw }PM\text{ perpendicular from }P\text{ to the directrix.}$
$\displaystyle \text{By the definition of a parabola, }SP=PM.$
$\displaystyle \therefore SP^2=PM^2$
$\displaystyle x^2+y^2=\left(\frac{2x-y-10}{\sqrt{2^2+(-1)^2}}\right)^2$
$\displaystyle x^2+y^2=\frac{(2x-y-10)^2}{5}$
$\displaystyle 5x^2+5y^2=(2x-y-10)^2$
$\displaystyle 5x^2+5y^2=4x^2+y^2+100-4xy-40x+20y$
$\displaystyle x^2+4y^2+4xy+40x-20y-100=0$
$\displaystyle \therefore \text{The required equation of the parabola is}$
$\displaystyle x^2+4y^2+4xy+40x-20y-100=0.$
$\displaystyle \\$

$\displaystyle \text{(ii) The given equation is }4x^2+y=0.$
$\displaystyle \Rightarrow 4x^2=-y$
$\displaystyle \Rightarrow x^2=-\frac{1}{4}y.$
$\displaystyle \text{Comparing it with }x^2=-4ay,$
$\displaystyle 4a=\frac{1}{4}$
$\displaystyle a=\frac{1}{16}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: }(0,0)$
$\displaystyle \text{Focus: }(0,-a)=\left(0,-\frac{1}{16}\right)$
$\displaystyle \text{Axis: }x=0$
$\displaystyle \text{Directrix: }y=a=\frac{1}{16}$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{1}{16}$
$\displaystyle =\frac{1}{4}\text{ unit}.$
$\displaystyle \\$

$\displaystyle \text{(iii) The given equation is }y^2-4y-3x+1=0.$
$\displaystyle \Rightarrow y^2-4y=3x-1$
$\displaystyle \Rightarrow y^2-4y+4=3x-1+4$
$\displaystyle \Rightarrow (y-2)^2=3x+3$
$\displaystyle \Rightarrow (y-2)^2=3(x+1) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(-1,2)\text{ without rotating the axes, let}$
$\displaystyle X=x+1\text{ and }Y=y-2. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) becomes}$
$\displaystyle Y^2=3X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=4aX,$
$\displaystyle 4a=3$
$\displaystyle a=\frac{3}{4}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(-1,2).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }\left(\frac{3}{4},0\right).$
$\displaystyle X=\frac{3}{4}\Rightarrow x+1=\frac{3}{4}$
$\displaystyle \Rightarrow x=-\frac{1}{4}.$
$\displaystyle Y=0\Rightarrow y-2=0$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \therefore \text{The focus is }\left(-\frac{1}{4},2\right).$
$\displaystyle \text{Directrix: }X=-a$
$\displaystyle \Rightarrow x+1=-\frac{3}{4}$
$\displaystyle \Rightarrow x=-\frac{7}{4}.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y-2=0$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{3}{4}$
$\displaystyle =3\text{ units}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the vertex, focus, axis, directrix and latus-rectum}$
$\displaystyle \text{of the following parabolas:}$
$\displaystyle \text{(iv) }y^2-4y+4x=0$
$\displaystyle \text{(v) }y^2+4x+4y-3=0$
$\displaystyle \text{(vi) }y^2=8x+8y$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(iv) The given equation is }y^2-4y+4x=0.$
$\displaystyle \Rightarrow y^2-4y=-4x$
$\displaystyle \Rightarrow y^2-4y+4=-4x+4$
$\displaystyle \Rightarrow (y-2)^2=-4(x-1) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(1,2)\text{ without rotating the axes, let}$
$\displaystyle X=x-1\text{ and }Y=y-2. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle Y^2=-4X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=-4aX,$
$\displaystyle 4a=4$
$\displaystyle a=1.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(1,2).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }(-a,0)=(-1,0).$
$\displaystyle X=-1\Rightarrow x-1=-1$
$\displaystyle \Rightarrow x=0.$
$\displaystyle Y=0\Rightarrow y-2=0$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \therefore \text{The focus is }(0,2).$
$\displaystyle \text{Directrix: }X=a$
$\displaystyle \Rightarrow x-1=1$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y-2=0$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times1$
$\displaystyle =4\text{ units}.$
$\displaystyle \\$

$\displaystyle \text{(v) The given equation is }y^2+4x+4y-3=0.$
$\displaystyle \Rightarrow y^2+4y=-4x+3$
$\displaystyle \Rightarrow y^2+4y+4=-4x+7$
$\displaystyle \Rightarrow (y+2)^2=-4x+7$
$\displaystyle \Rightarrow (y+2)^2=-4\left(x-\frac{7}{4}\right) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }\left(\frac{7}{4},-2\right)\text{ without rotating the axes, let}$
$\displaystyle X=x-\frac{7}{4}\text{ and }Y=y+2. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle Y^2=-4X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=-4aX,$
$\displaystyle 4a=4$
$\displaystyle a=1.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }\left(\frac{7}{4},-2\right).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }(-a,0)=(-1,0).$
$\displaystyle X=-1\Rightarrow x-\frac{7}{4}=-1$
$\displaystyle \Rightarrow x=\frac{3}{4}.$
$\displaystyle Y=0\Rightarrow y+2=0$
$\displaystyle \Rightarrow y=-2.$
$\displaystyle \therefore \text{The focus is }\left(\frac{3}{4},-2\right).$
$\displaystyle \text{Directrix: }X=a$
$\displaystyle \Rightarrow x-\frac{7}{4}=1$
$\displaystyle \Rightarrow x=\frac{11}{4}.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y+2=0$
$\displaystyle \Rightarrow y=-2.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times1$
$\displaystyle =4\text{ units}.$
$\displaystyle \\$

$\displaystyle \text{(vi) The given equation is }y^2=8x+8y.$
$\displaystyle \Rightarrow y^2-8y=8x$
$\displaystyle \Rightarrow y^2-8y+16=8x+16$
$\displaystyle \Rightarrow (y-4)^2=8(x+2) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(-2,4)\text{ without rotating the axes, let}$
$\displaystyle X=x+2\text{ and }Y=y-4. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle Y^2=8X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=4aX,$
$\displaystyle 4a=8$
$\displaystyle a=2.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(-2,4).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }(a,0)=(2,0).$
$\displaystyle X=2\Rightarrow x+2=2$
$\displaystyle \Rightarrow x=0.$
$\displaystyle Y=0\Rightarrow y-4=0$
$\displaystyle \Rightarrow y=4.$
$\displaystyle \therefore \text{The focus is }(0,4).$
$\displaystyle \text{Directrix: }X=-a$
$\displaystyle \Rightarrow x+2=-2$
$\displaystyle \Rightarrow x=-4.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y-4=0$
$\displaystyle \Rightarrow y=4.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times2$
$\displaystyle =8\text{ units}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find the vertex, focus, axis, directrix and latus-rectum}$
$\displaystyle \text{of the following parabolas:}$
$\displaystyle \text{(vii) }4(y-1)^2=-7(x-3)$
$\displaystyle \text{(viii) }y^2=5x-4y-9$
$\displaystyle \text{(ix) }x^2+y=6x-14$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(vii) The given equation is }4(y-1)^2=-7(x-3).$
$\displaystyle \Rightarrow (y-1)^2=-\frac{7}{4}(x-3) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(3,1)\text{ without rotating the axes, let}$
$\displaystyle X=x-3\text{ and }Y=y-1. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle Y^2=-\frac{7}{4}X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=-4aX,$
$\displaystyle 4a=\frac{7}{4}$
$\displaystyle a=\frac{7}{16}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(3,1).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }\left(-\frac{7}{16},0\right).$
$\displaystyle X=-\frac{7}{16}\Rightarrow x-3=-\frac{7}{16}$
$\displaystyle \Rightarrow x=\frac{41}{16}.$
$\displaystyle Y=0\Rightarrow y-1=0$
$\displaystyle \Rightarrow y=1.$
$\displaystyle \therefore \text{The focus is }\left(\frac{41}{16},1\right).$
$\displaystyle \text{Directrix: }X=a$
$\displaystyle \Rightarrow x-3=\frac{7}{16}$
$\displaystyle \Rightarrow x=\frac{55}{16}.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y-1=0$
$\displaystyle \Rightarrow y=1.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{7}{16}$
$\displaystyle =\frac{7}{4}\text{ units}.$
$\displaystyle \\$

$\displaystyle \text{(viii) The given equation is }y^2=5x-4y-9.$
$\displaystyle \Rightarrow y^2+4y=5x-9$
$\displaystyle \Rightarrow y^2+4y+4=5x-9+4$
$\displaystyle \Rightarrow (y+2)^2=5x-5$
$\displaystyle \Rightarrow (y+2)^2=5(x-1) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(1,-2)\text{ without rotating the axes, let}$
$\displaystyle X=x-1\text{ and }Y=y+2. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle Y^2=5X. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }Y^2=4aX,$
$\displaystyle 4a=5$
$\displaystyle a=\frac{5}{4}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(1,-2).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }\left(\frac{5}{4},0\right).$
$\displaystyle X=\frac{5}{4}\Rightarrow x-1=\frac{5}{4}$
$\displaystyle \Rightarrow x=\frac{9}{4}.$
$\displaystyle Y=0\Rightarrow y+2=0$
$\displaystyle \Rightarrow y=-2.$
$\displaystyle \therefore \text{The focus is }\left(\frac{9}{4},-2\right).$
$\displaystyle \text{Directrix: }X=-a$
$\displaystyle \Rightarrow x-1=-\frac{5}{4}$
$\displaystyle \Rightarrow x=-\frac{1}{4}.$
$\displaystyle \text{Axis: }Y=0$
$\displaystyle \Rightarrow y+2=0$
$\displaystyle \Rightarrow y=-2.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{5}{4}$
$\displaystyle =5\text{ units}.$
$\displaystyle \\$

$\displaystyle \text{(ix) The given equation is }x^2+y=6x-14.$
$\displaystyle \Rightarrow x^2-6x=-y-14$
$\displaystyle \Rightarrow x^2-6x+9=-y-14+9$
$\displaystyle \Rightarrow (x-3)^2=-(y+5) \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \text{Shifting the origin to }(3,-5)\text{ without rotating the axes, let}$
$\displaystyle X=x-3\text{ and }Y=y+5. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Using these relations, equation (i) reduces to}$
$\displaystyle X^2=-Y. \hspace{0.5cm}\text{... ... ... (iii)}$
$\displaystyle \text{Comparing it with }X^2=-4aY,$
$\displaystyle 4a=1$
$\displaystyle a=\frac{1}{4}.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Vertex: In the new coordinate system, the vertex is }(0,0).$
$\displaystyle \text{Hence, in the original coordinate system, the vertex is }(3,-5).$
$\displaystyle \text{Focus: In the new coordinate system, the focus is }\left(0,-\frac{1}{4}\right).$
$\displaystyle X=0\Rightarrow x-3=0$
$\displaystyle \Rightarrow x=3.$
$\displaystyle Y=-\frac{1}{4}\Rightarrow y+5=-\frac{1}{4}$
$\displaystyle \Rightarrow y=-\frac{21}{4}.$
$\displaystyle \therefore \text{The focus is }\left(3,-\frac{21}{4}\right).$
$\displaystyle \text{Directrix: }Y=a$
$\displaystyle \Rightarrow y+5=\frac{1}{4}$
$\displaystyle \Rightarrow y=-\frac{19}{4}.$
$\displaystyle \text{Axis: }X=0$
$\displaystyle \Rightarrow x-3=0$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{1}{4}$
$\displaystyle =1\text{ unit}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{For the parabola }y^2=4px\text{, find the extremities}$
$\displaystyle \text{of a double ordinate of length }8p.\text{ Prove that the lines from the vertex}$
$\displaystyle \text{to its extremities are at right angles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }PQ\text{ be a double ordinate of length }8p\text{ of the parabola }y^2=4px.$
$\displaystyle \text{Since the axis bisects the double ordinate, let it meet }PQ\text{ at }R.$
$\displaystyle \therefore PR=QR=4p.$
$\displaystyle \text{Let }AR=x_1,\text{ where }A(0,0)\text{ is the vertex.}$
$\displaystyle \therefore P=(x_1,4p)\text{ and }Q=(x_1,-4p).$
$\displaystyle \text{Since }P\text{ lies on }y^2=4px,$
$\displaystyle (4p)^2=4px_1$
$\displaystyle 16p^2=4px_1$
$\displaystyle x_1=4p.$
$\displaystyle \therefore P=(4p,4p)\text{ and }Q=(4p,-4p).$
$\displaystyle \text{Now, the slope of }AP\text{ is}$
$\displaystyle m_1=\frac{4p-0}{4p-0}=1.$
$\displaystyle \text{The slope of }AQ\text{ is}$
$\displaystyle m_2=\frac{-4p-0}{4p-0}=-1.$
$\displaystyle \therefore m_1m_2=(1)(-1)=-1.$
$\displaystyle \therefore AP\perp AQ.$
$\displaystyle \text{Hence, the lines from the vertex to the extremities of the double ordinate}$
$\displaystyle \text{are at right angles.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Find the area of the triangle formed by the lines joining}$
$\displaystyle \text{the vertex of the parabola }x^2=12y\text{ to the ends of its latus-rectum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation of the parabola is }x^2=12y.$
$\displaystyle \text{Comparing it with }x^2=4ay,$
$\displaystyle 4a=12$
$\displaystyle a=3.$
$\displaystyle \text{Therefore, the vertex is }O=(0,0)\text{ and the focus is }S=(0,3).$
$\displaystyle \text{Let }P\text{ and }Q\text{ be the ends of the latus-rectum.}$
$\displaystyle \text{Since the latus-rectum passes through the focus, its equation is }y=3.$
$\displaystyle \text{Substituting }y=3\text{ in }x^2=12y,$
$\displaystyle x^2=12\times3$
$\displaystyle x^2=36$
$\displaystyle x=\pm6.$
$\displaystyle \therefore P=(-6,3)\text{ and }Q=(6,3).$
$\displaystyle PQ=\sqrt{(6+6)^2+(3-3)^2}$
$\displaystyle =\sqrt{12^2}$
$\displaystyle =12.$
$\displaystyle \text{The perpendicular distance from }O\text{ to }PQ\text{ is }OS=3.$
$\displaystyle \therefore \text{Area of }\triangle POQ=\frac{1}{2}\times PQ\times OS$
$\displaystyle =\frac{1}{2}\times12\times3$
$\displaystyle =18\text{ square units}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find the coordinates of the point of intersection of the axis}$
$\displaystyle \text{and the directrix of the parabola whose focus is }(3,3)\text{ and directrix is}$
$\displaystyle 3x-4y=2.\text{ Find also the length of its latus-rectum.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The equation of the directrix is}$
$\displaystyle 3x-4y=2. \hspace{0.5cm}\text{... ... ... (i)}$
$\displaystyle \Rightarrow y=\frac{3}{4}x-\frac{1}{2}$
$\displaystyle \therefore \text{Slope of the directrix}=\frac{3}{4}.$
$\displaystyle \text{The axis is perpendicular to the directrix.}$
$\displaystyle \therefore \text{Slope of the axis}=-\frac{1}{\frac{3}{4}}=-\frac{4}{3}.$
$\displaystyle \text{Since the focus }(3,3)\text{ lies on the axis, its equation is}$
$\displaystyle y-3=-\frac{4}{3}(x-3)$
$\displaystyle \Rightarrow 3y-9=-4x+12$
$\displaystyle \Rightarrow 4x+3y=21. \hspace{0.5cm}\text{... ... ... (ii)}$
$\displaystyle \text{Solving equations (i) and (ii), we get}$
$\displaystyle 3x-4y=2$
$\displaystyle 4x+3y=21.$
$\displaystyle \text{Multiplying the first equation by }3\text{ and the second equation by }4,$
$\displaystyle 9x-12y=6$
$\displaystyle 16x+12y=84.$
$\displaystyle \text{Adding, we get}$
$\displaystyle 25x=90$
$\displaystyle x=\frac{18}{5}.$
$\displaystyle \text{Substituting }x=\frac{18}{5}\text{ in }4x+3y=21,$
$\displaystyle \frac{72}{5}+3y=21$
$\displaystyle 3y=\frac{33}{5}$
$\displaystyle y=\frac{11}{5}.$
$\displaystyle \therefore \text{The point of intersection of the axis and directrix is}$
$\displaystyle \left(\frac{18}{5},\frac{11}{5}\right).$
$\displaystyle \text{The perpendicular distance of the focus }(3,3)\text{ from the directrix is}$
$\displaystyle \frac{|3(3)-4(3)-2|}{\sqrt{3^2+(-4)^2}}$
$\displaystyle =\frac{|-5|}{5}$
$\displaystyle =1.$
$\displaystyle \text{For a parabola, the distance of the focus from the directrix is }2a.$
$\displaystyle \therefore 2a=1$
$\displaystyle a=\frac{1}{2}.$
$\displaystyle \therefore \text{Length of the latus-rectum}=4a$
$\displaystyle =4\times\frac{1}{2}$
$\displaystyle =2\text{ units}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{At what point of the parabola }x^2=9y\text{ is the abscissa}$
$\displaystyle \text{three times the ordinate?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the ordinate of the required point be }y.$
$\displaystyle \text{Then the abscissa is }3y.$
$\displaystyle \therefore \text{The coordinates of the required point are }(3y,y).$
$\displaystyle \text{Since this point lies on the parabola }x^2=9y,$
$\displaystyle (3y)^2=9y$
$\displaystyle 9y^2=9y$
$\displaystyle y(y-1)=0.$
$\displaystyle \text{Ignoring the trivial solution }y=0,\text{ we get }y=1.$
$\displaystyle \therefore x=3y=3.$
$\displaystyle \therefore \text{The required point is }(3,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Find the equation of a parabola with vertex at the origin,}$
$\displaystyle \text{the axis along the x-axis and passing through }(2,3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since the axis is along the positive x-axis, let the equation of the parabola be}$
$\displaystyle y^2=4ax.$
$\displaystyle \text{As }(2,3)\text{ lies on the parabola,}$
$\displaystyle 3^2=4a(2)$
$\displaystyle 9=8a$
$\displaystyle a=\frac{9}{8}.$
$\displaystyle \therefore y^2=4\times\frac{9}{8}x$
$\displaystyle y^2=\frac{9}{2}x$
$\displaystyle \therefore 2y^2=9x.$
$\displaystyle \text{Hence, the required equation of the parabola is }2y^2=9x.$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Find the equation of the parabola with vertex at the origin}$
$\displaystyle \text{and directrix }y=2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }Z(x_1,y_1)\text{ be the point of intersection of the axis and the directrix.}$
$\displaystyle \text{Since the directrix is }y=2\text{ and the axis is perpendicular to it through the origin,}$
$\displaystyle Z=(0,2).$
$\displaystyle \text{Let the focus be }S(x_2,y_2).$
$\displaystyle \text{The vertex }O(0,0)\text{ is the midpoint of }SZ.$
$\displaystyle \therefore \frac{x_2+0}{2}=0\text{ and }\frac{y_2+2}{2}=0$
$\displaystyle \Rightarrow x_2=0\text{ and }y_2=-2.$
$\displaystyle \therefore \text{The focus is }S=(0,-2).$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the parabola.}$
$\displaystyle \text{By the definition of a parabola,}$
$\displaystyle PS=PM,$
$\displaystyle \text{where }PM\text{ is the perpendicular distance of }P\text{ from the directrix }y=2.$
$\displaystyle \therefore PS^2=PM^2$
$\displaystyle (x-0)^2+(y+2)^2=(y-2)^2$
$\displaystyle \Rightarrow x^2+y^2+4y+4=y^2-4y+4$
$\displaystyle \Rightarrow x^2=-8y.$
$\displaystyle \therefore x^2+8y=0.$
$\displaystyle \text{Hence, the required equation of the parabola is }x^2+8y=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Find the equation of the parabola whose focus is }(5,2)$
$\displaystyle \text{and whose vertex is }(3,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=(3,2)\text{ be the vertex and }S=(5,2)\text{ be the focus.}$
$\displaystyle \text{Since the vertex and focus have the same }y\text{-coordinate, the axis is horizontal.}$
$\displaystyle \text{The parabola opens towards the positive }x\text{-axis.}$
$\displaystyle a=AS$
$\displaystyle =5-3$
$\displaystyle =2.$
$\displaystyle \text{The standard equation of a parabola with vertex }(h,k)\text{ and horizontal axis is}$
$\displaystyle (y-k)^2=4a(x-h).$
$\displaystyle \text{Here, }h=3,\ k=2\text{ and }a=2.$
$\displaystyle \therefore (y-2)^2=4(2)(x-3)$
$\displaystyle (y-2)^2=8(x-3).$
$\displaystyle \Rightarrow y^2-4y+4=8x-24$
$\displaystyle \Rightarrow y^2-4y-8x+28=0.$
$\displaystyle \text{Hence, the required equation of the parabola is}$
$\displaystyle y^2-4y-8x+28=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{The cable of a uniformly loaded suspension bridge hangs}$
$\displaystyle \text{in the form of a parabola. The horizontal roadway is }100\text{ m long and is}$
$\displaystyle \text{supported by vertical wires attached to the cable. The longest wire is }30\text{ m}$
$\displaystyle \text{and the shortest wire is }6\text{ m. Find the length of a supporting wire attached}$
$\displaystyle \text{to the roadway }18\text{ m from the middle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the vertex of the parabola be the origin and let its axis be the positive }y\text{-axis.}$
$\displaystyle \text{The shortest wire is at the middle of the roadway and has length }6\text{ m}.$
$\displaystyle \text{Therefore, the roadway lies }6\text{ m above the vertex of the parabola.}$
$\displaystyle \text{Let }AB\text{ be the longest wire and }OC\text{ be the shortest wire.}$
$\displaystyle \text{Let }DF\text{ be the supporting wire situated }18\text{ m from the middle.}$
$\displaystyle AB=30\text{ m},\quad OC=6\text{ m}$
$\displaystyle BC=\frac{100}{2}=50\text{ m}.$
$\displaystyle \text{Since the parabola opens upwards, let its equation be}$
$\displaystyle x^2=4ay.$
$\displaystyle \text{At the end of the roadway, the height of the cable above the vertex is}$
$\displaystyle 30-6=24\text{ m}.$
$\displaystyle \therefore A=(50,24).$
$\displaystyle \text{Since }A\text{ lies on the parabola,}$
$\displaystyle 50^2=4a(24)$
$\displaystyle 2500=96a$
$\displaystyle a=\frac{625}{24}.$
$\displaystyle \therefore x^2=4\left(\frac{625}{24}\right)y$
$\displaystyle x^2=\frac{625}{6}y$
$\displaystyle \Rightarrow 6x^2=625y.$
$\displaystyle \text{For the supporting wire }18\text{ m from the middle, }x=18.$
$\displaystyle 6(18)^2=625y$
$\displaystyle y=\frac{6\times18^2}{625}$
$\displaystyle =\frac{1944}{625}$
$\displaystyle =3.1104\text{ m}.$
$\displaystyle \text{Therefore, the length of the supporting wire is}$
$\displaystyle DF=6+y$
$\displaystyle =6+\frac{1944}{625}$
$\displaystyle =\frac{5694}{625}$
$\displaystyle =9.1104\text{ m}.$
$\displaystyle \therefore \text{The required length of the supporting wire is approximately }9.11\text{ m}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Find the equations of the lines joining the vertex of the}$
$\displaystyle \text{parabola }y^2=6x\text{ to the points on it whose abscissa is }24.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A\text{ and }B\text{ be the points on the parabola }y^2=6x$
$\displaystyle \text{whose abscissa is }24,\text{ and let }O\text{ be the vertex.}$
$\displaystyle \text{When }x=24,$
$\displaystyle y^2=6\times24=144$
$\displaystyle \Rightarrow y=\pm12.$
$\displaystyle \therefore A=(24,12)\text{ and }B=(24,-12).$
$\displaystyle \text{The required lines pass through the origin }(0,0).$
$\displaystyle \text{Their equations are}$
$\displaystyle y=\frac{12}{24}x=\frac{x}{2}$
$\displaystyle \Rightarrow x-2y=0,$
$\displaystyle \text{and}$
$\displaystyle y=\frac{-12}{24}x=-\frac{x}{2}$
$\displaystyle \Rightarrow x+2y=0.$
$\displaystyle \therefore \text{The required equations are }x-2y=0\text{ and }x+2y=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Find the coordinates of the points on the parabola }y^2=8x$
$\displaystyle \text{whose focal distance is }4.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given parabola }y^2=8x.$
$\displaystyle \Rightarrow y^2=4(2)x.$
$\displaystyle \text{Comparing with the standard equation }y^2=4ax,\text{ we get }a=2.$
$\displaystyle \text{Let }(x_1,y_1)\text{ be the required point.}$
$\displaystyle \text{The focal distance of a point on }y^2=4ax\text{ is }x+a.$
$\displaystyle \text{Given focal distance }=4,$
$\displaystyle x_1+a=4$
$\displaystyle \Rightarrow x_1+2=4$
$\displaystyle \Rightarrow x_1=2.$
$\displaystyle \text{Since }(x_1,y_1)\text{ lies on the parabola,}$
$\displaystyle y_1^2=8(2)=16$
$\displaystyle \Rightarrow y_1=\pm4.$
$\displaystyle \therefore \text{The required points are }(2,4)\text{ and }(2,-4).$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Find the length of the line segment joining the vertex of the}$
$\displaystyle \text{parabola }y^2=4ax\text{ to a point on it, if the line segment makes an angle }\theta$
$\displaystyle \text{with the positive x-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be the vertex and }B(x_1,y_1)\text{ be the point on the parabola.}$
$\displaystyle \text{Let }A\text{ be the foot of the perpendicular from }B\text{ to the x-axis.}$
$\displaystyle \text{Suppose }OB\text{ makes an angle }\theta\text{ with the positive x-axis.}$
$\displaystyle \text{In the right-angled triangle }OAB,$
$\displaystyle \cos\theta=\frac{OA}{OB}=\frac{x_1}{OB}$
$\displaystyle \text{and}\quad \sin\theta=\frac{AB}{OB}=\frac{y_1}{OB}.$
$\displaystyle \therefore x_1=OB\cos\theta$
$\displaystyle \text{and}\quad y_1=OB\sin\theta.$
$\displaystyle \text{Since }B(x_1,y_1)\text{ lies on the parabola }y^2=4ax,$
$\displaystyle y_1^2=4ax_1.$
$\displaystyle \therefore (OB\sin\theta)^2=4a(OB\cos\theta)$
$\displaystyle OB^2\sin^2\theta=4a\,OB\cos\theta$
$\displaystyle \Rightarrow OB\sin^2\theta=4a\cos\theta$
$\displaystyle \Rightarrow OB=\frac{4a\cos\theta}{\sin^2\theta}$
$\displaystyle =4a\,\mathrm{cosec}\,\theta\cot\theta.$
$\displaystyle \therefore \text{The required length is }4a\,\mathrm{cosec}\,\theta\cot\theta.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If the points }(0,4)\text{ and }(0,2)\text{ are respectively the vertex}$
$\displaystyle \text{and focus of a parabola, find its equation.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }A=(0,4)\text{ be the vertex and }F=(0,2)\text{ be the focus.}$
$\displaystyle \text{Since the vertex and focus lie on the y-axis, the y-axis is the axis of the parabola.}$
$\displaystyle \text{Let the directrix meet the axis at }Z.$
$\displaystyle \text{Since the vertex is the midpoint of }FZ,$
$\displaystyle AZ=AF=4-2=2.$
$\displaystyle \therefore Z=(0,6).$
$\displaystyle \text{Hence, the equation of the directrix is}$
$\displaystyle y=6.$
$\displaystyle \text{Let }P(x,y)\text{ be any point on the parabola and let }PM\text{ be its perpendicular}$
$\displaystyle \text{distance from the directrix. By the definition of a parabola,}$
$\displaystyle PF=PM.$
$\displaystyle \therefore \sqrt{(x-0)^2+(y-2)^2}=|y-6|$
$\displaystyle \Rightarrow x^2+(y-2)^2=(y-6)^2$
$\displaystyle \Rightarrow x^2+y^2-4y+4=y^2-12y+36$
$\displaystyle \Rightarrow x^2+8y=32.$
$\displaystyle \therefore \text{The required equation of the parabola is }x^2+8y=32.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If the line }y=mx+1\text{ is tangent to the parabola }y^2=4x,$
$\displaystyle \text{find the value of }m.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given parabola }y^2=4x\text{ and line }y=mx+1.$
$\displaystyle \text{Substituting }y=mx+1\text{ in }y^2=4x,\text{ we get}$
$\displaystyle (mx+1)^2=4x$
$\displaystyle \Rightarrow m^2x^2+2mx+1=4x$
$\displaystyle \Rightarrow m^2x^2+(2m-4)x+1=0.$
$\displaystyle \text{Since the line is tangent to the parabola, the roots of this quadratic are equal.}$
$\displaystyle \therefore D=0$
$\displaystyle (2m-4)^2-4(m^2)(1)=0$
$\displaystyle \Rightarrow 4m^2-16m+16-4m^2=0$
$\displaystyle \Rightarrow -16m+16=0$
$\displaystyle \Rightarrow m=1.$
$\displaystyle \therefore \text{The required value of }m\text{ is }1.$
$\displaystyle \\$